Hc Verma I for Class 11 Science Physics Chapter 18 - Geometrical Optics

Download PdfDownload Pdf

Share with your friends

Hc Verma I Solutions for Class 11 Science Physics Chapter 18 Geometrical Optics are provided here with simple step-by-step explanations. These solutions for Geometrical Optics are extremely popular among Class 11 Science students for Physics Geometrical Optics Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Hc Verma I Book of Class 11 Science Physics Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Hc Verma I Solutions. All Hc Verma I Solutions for class Class 11 Science Physics are prepared by experts and are 100% accurate.

Page No 410:

Answer:

No, because this formula is only valid when the angles of incidence and refraction are small. When the angles of incidence and refraction are large, we cannot use relations sini = tani and sinr = tanr to derive the formula for the apparent depth.

Page No 410:

Answer:

There is only one such situation. Light rays go undeviated through a prism only when the prism is kept in a medium that has the same refractive index as that of the prism itself. In all other situations, there will always be some minimum deviation in the path of a light ray passing through the prism.

Page No 410:

Answer:

A diamond shines more because of the phenomenon of total internal reflection (TIR). The refractive index of diamond is ≈2.4 and that of glass is ≈1.5.
By using the relation $μ = \frac{1}{\sin C}$ , where C is the critical angle, we get refractive index (μ).
For a large refractive index, the value of the critical angle is small. Thus, for the diamond, the critical angle for total internal reflection is much smaller than that of the glass. Therefore, a great percentage of incident light gets internally reflected several times before it emerges out.

Page No 410:

Answer:

The beam of light will seem to appear and disappear, just like the twinkling of the stars. This is because the refractive index of the material in the slab fluctuates slowly with time. So, the light rays get refracted differently with time and this causes them to get concentrated during certain times, or get diffused at other times.

Page No 410:

Answer:

No, a plane mirror can never form a real image. This is because in all possible situations of image formation, the light rays never actually meet after getting reflected, but they only appear to meet behind the mirror, forming a virtual image always.

Page No 410:

Answer:

No, the paper will not burn even after sufficient time. This is because the piece of paper is placed at the position of a virtual image of a strong light source, so light rays are not actually converging at that point, but seem to converge. As no light rays are concentrating on the point, the paper will never burn.
Instead, if the paper is placed at the position of a real image, it will burn as the light rays are actually concentrating on the point. Also, when we have the real image of a virtual source, it will also cause the burning of paper due to same reason; the light rays are actually meeting at the point.

Page No 410:

Answer:

Yes, a virtual image can be photographed by a camera. This is because the light rays emitting from a virtual image and reaching the lens of the camera are real. A virtual image is formed from a reflecting surface after reflection of real incident rays, so these rays enter the camera to provide effects on the photographic film. Just like a virtual image can be seen by us in a mirror with our eyes, it can be photographed by a camera.

Page No 410:

Answer:

The special function of a convex mirror is that it creates the image of a distant object that is reduced in size, is upright or erect and always lies within the virtual focal length of the mirror. A plane mirror cannot do this. Also, as the image is formed within the focal length, the image is close to the mirror as well as is small in size, enabling the driver to clearly view the nearer vehicles behind the motor vehicle.

Page No 410:

Answer:

The image of the object moves slower compared to the object. It can be explained using the mirror formula : $\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
We know that for a convex mirror, the object distance (u) is positive, image distance (v) is negative and the focal length (f) is also negative. Thus mirror formula of a convex mirror is: $\frac{1}{u} - \frac{1}{v} = - \frac{1}{f}$
As u = +ve
$\frac{1}{v} - \frac{1}{f} > 0 \frac{1}{v} > \frac{1}{f} v < f$
Therefore, the image is always formed within the focal length of the mirror. Thus, the distance moved by the image is much slower than the distance moved by the object.

Page No 410:

Answer:

When viewed from the water, the friend will seem taller than his usual height.

Let actual height be h and the apparent height be h'.
Here, the refraction is taking place from rarer to denser medium and a virtual image is formed.
Using
$\frac{μ_{1}}{- u} + \frac{μ_{2}}{v} = \frac{μ_{2} - μ_{1}}{R}$
Where refractive index of water is μ₂ and refractive index of air is μ₁.
u and v are object and image distances, respectively.
R is the radius of curvature, here we will take it as ∞.
$\frac{μ_{1}}{- u} + \frac{μ_{2}}{v} = \frac{μ_{2} - μ_{1}}{\infty} \frac{μ_{1}}{u} = \frac{μ_{2}}{v} v = \frac{μ_{2}}{μ_{1}} \times u$
As μ₁= 1
v = u × μ₂
We know magnification is given by:
$m = \frac{v}{u}$
Putting the value of v in the above equation:
$m = \frac{u \times μ_{2}}{u} m = μ_{2}$
As the magnification is greater than 1, so the apparent height seems to be greater than actual height.

Page No 410:

Answer:

Proof:
$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R} Now R = \infty \frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{\infty} \frac{μ_{2}}{v} - \frac{μ_{1}}{u} = 0 \frac{μ_{2}}{v} = \frac{μ_{1}}{u} \frac{μ_{1}}{μ_{2}} = \frac{u}{v} But \frac{u}{v} = \frac{Real depth / height}{Apparent depth / height} ∴ \frac{Real depth / height}{Apparent depth / height} = \frac{μ_{1}}{μ_{2}}$

Page No 410:

Answer:

Yes. Using the lens maker formula we can show it. We know that the formula is:
$\frac{1}{f} = (\frac{μ_{2}}{μ_{1}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Where, f is the focal length of the thin converging lens.
μ₂ and μ₁ are refractive indexes of lens and air respectively.
R₁ and R₂ are the radius of curvature of convex and plane surfaces respectively.
Here, R₂= ∞ because of the plane surface.

Case 1
When the convex surface is facing the object, we have:
$\frac{1}{f} = (\frac{μ_{2}}{μ_{1}} - 1) (\frac{1}{R_{1}} - \frac{1}{\infty}) \frac{1}{f} = (\frac{μ_{2}}{μ_{1}} - 1) \frac{1}{R_{1}} . . . (i)$

Case 2

When the plane surface is facing the object, we have:
$\frac{1}{f} = (\frac{μ_{2}}{μ_{1}} - 1) (\frac{1}{\infty} - \frac{1}{R_{1}}) \frac{1}{f} = - (\frac{μ_{2}}{μ_{1}} - 1) \frac{1}{R_{1}} . . . (ii)$
For Case 1, the focal length is positive and for the Case 2 the focal length is negative. Thus, the image distance is different in both cases.

Page No 410:

Answer:

Yes, we can say so with certainty, as only a diverging lens can cause a parallel beam of light to diverge.

Page No 410:

Answer:

It will act as a diverging lens. This is because the refractive index of air is less than that of water. Thus, light will diverge after passing through the bubble.

Page No 410:

Answer:

Let the two converging lenses be L₁ and L₂, with focal lengths f₁ and f₂ respectively.
To reduce the aperture of a parallel beam of light without losing the energy of the light and also increase the intensity, we have to place the lens L₂ within the focal range of L_1.

Page No 410:

Answer:

No, the focal length of mirror will not change. This is because the focal length of a mirror does not depend on the refractive index of the medium in which the light rays travel.

Page No 410:

Answer:

Yes, the focal length of the lens will change. This is because its focal length is dependent on the medium in which the light rays travel.

Page No 410:

Answer:

No, mirrors cannot give rise to chromatic aberration. This is because chromatic aberration occurs due to the refraction of different colours of light. In case of mirrors, refraction of light does not take place.

Page No 410:

Answer:

No, there will not be a significant chromatic aberration because a laser light is monochromatic in nature. Therefore, all light will get converged at the same point.

Page No 410:

Answer:

(a) All the reflected rays meet at a point when produced backward.

Here, the angle of reflection is equal to the angle of incidence. Therefore, all rays get reflected to converge at a single point to form the point image of the point source.

Page No 411:

Answer:

(b) light goes from optically denser medium to rarer medium

In this case the incident angle is greater than critical angle, so the light gets reflected back into the same denser medium, instead of being refracted .

Page No 411:

Answer:

(c) form nearly a point image of a point source

Since, when reflected back, they meet at a single point forming a point image of a point source.

Page No 411:

Answer:

(d) 15 cm behind the mirror

Since u = − 30 cm
v = ?
f = 30 cm

From the mirror formula:
$\frac{1}{u} + \frac{1}{v} = \frac{1}{f} or \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \Rightarrow \frac{1}{v} = \frac{1}{30} - \frac{1}{- 30} \frac{1}{v} = \frac{1}{30} + \frac{1}{30} \frac{1}{v} = \frac{2}{30} ∴ v = 15 cm$

Page No 411:

Answer:

(a) is plane

This is because the reflected rays are still parallel, which is only possible if the mirror is a plane mirror. A spherical mirror will either converge or diverge the reflected rays.

Page No 411:

Answer:

(c) is certainly real if the object is virtual

As a virtual object is the virtual image of the object, it is always formed beyond the focus of the concave mirror. Thus, the image formed by a concave mirror of the virtual object will always be real.

Page No 411:

Answer:

(d) two images form, one at O' and the other at O"

Light will be refracted differently in both mediums on the right side. Thus, two images will be formed, one at O' due to refraction from medium μ₁ and another at O" due to refraction from medium μ₃.μ1

Page No 411:

Answer:

(c) $\frac{1}{v - t} - \frac{1}{u + t} = \frac{1}{f}$

The thickness of lens will cause a shift in the position of object and image from the ideal condition, which will be:
$u_{new} = u + t (1 + \frac{1}{μ}) and v_{new} = v - t (1 + \frac{1}{μ}) o r u_{new} = u + t + \frac{t}{μ} and v_{new} = v - t + \frac{t}{μ}$
As $\frac{t}{μ}$ will be quite small, it can be ignored.
Thus, u_new = u + t and v_new = v − t
We get the new lens formula:
$\frac{1}{v - t} - \frac{1}{u + t} = \frac{1}{f}$

Page No 411:

Answer:

(b) f = R

As from lens maker formula:
$\frac{1}{f} = (m - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \Rightarrow \frac{1}{f} = (1.5 - 1) (\frac{1}{R} - \frac{1}{- R}) (As given R_{1} = R_{2} = R) \frac{1}{f} = (0.5) (\frac{1}{R} + \frac{1}{R}) \frac{1}{f} = (0.5) (\frac{2}{R}) f = R$ =R

Page No 411:

Answer:

(c) 2 f

Since the object is placed at 2 f, the image of the object will be formed at distance of 2 f from a converging lens.
It can also be shown from the lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Here, u = − 2 f and f = f
On putting the respective values we get:
$\frac{1}{v} - \frac{1}{- 2 f} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{1}{2 f} = \frac{1}{2 f}$
Therefore, image distance v = 2 f

Page No 411:

Answer:

(d) first increases then decreases

Since all beams of light are parallel to the principal axis, they will converge at the focus of a converging lens. Thus, the intensity of light increases till one reaches the focus and then starts decreasing as one moves beyond it.

Page No 411:

Answer:

(a) 2 D

The lens is cut into two equal parts by a plane perpendicular to the principal axis. Thus, the radius of curvature of both the lenses become half of its original value.
As,
$f = \frac{R}{2} P = \frac{1}{f}$
Radius of curvature becomes half. Therefore, the focal also reduces to half its original value.
Thus, the power (P) of the two cut-lenses will be equal.
2 P = 4 D
P = 2 D

Page No 411:

Answer:

(c) 4 D

As the lens is cut along the principal axis, the two new lens will act as two double convex lenses. Even after cutting the lens, the radius of the curvature of the two new lens will remain the same as that of the original lens, with the power equal to the original lens.

Page No 411:

Answer:

(c) increases

The focal length of the combination will increase. For finding out the combination of lens we have the formula:
$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{d}{f_{1} f_{2}}$
Where, F is the focal length for the combination
d is the separation between two lenses
Here, d = 0
$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} F = \frac{f_{1} f_{2}}{f_{1} + f_{2}}$
Hence, the focal length will increase.

Page No 411:

Answer:

(a) a convergent lens

Since the refractive index of lens is more than that of of water and both the sides are convex, it will act like a normal convergent lens (though its focal length will change).

Page No 411:

Answer:

(b) a divergent lens

Since the refractive index of lens is less than that of of water and both the sides are convex, it will act like a normal diverging lens (though its focal length will change to negative).

Page No 411:

Answer:

(b) 2.5 cm

As the focal length of the lens is 20 cm and object distance is 40 cm from the lens, the image is formed at the centre of curvature at the right side of the lens.

From right angled triangle ABC,
$\tan α = \frac{AB}{BC} \tan α = \frac{5}{40} α = \tan^{- 1} (\frac{5}{40})$
and from right angled triangle, we have:
$\tan α = \frac{DE}{DC} \tan α = \frac{h}{20} as DB - BC = DC = 20 cm$

Putting the value of angle alpha, we get:
$\tan (\tan^{- 1} (\frac{5}{40})) = \frac{h}{20} \frac{5}{40} = \frac{h}{20} \Rightarrow h = 2.5 cm$

Page No 411:

Answer:

(d) chromatic aberration

When light rays of different colours do not converge at the same point after passing through a converging lens, it is called chromatic aberration. This happens because a lens has different refractive indices for different colours, i.e, for different wavelengths of light.

Page No 412:

Answer:

Using sign conventions, given,
Distance of object from mirror, u = − 30 cm,
Radius of curvature of concave mirror R = − 40 cm

Using the mirror equation,

$\frac{1}{v} + \frac{1}{u} = \frac{2}{R} \Rightarrow \frac{1}{v} = \frac{2}{R} - \frac{1}{u} \Rightarrow \frac{1}{v} = \frac{2}{- 40} - \frac{1}{- 30} = \frac{1}{- 20} + \frac{1}{30} \Rightarrow \frac{1}{v} = \frac{- 30 + 20}{30 \times 20} = \frac{- 10}{30 \times 20} \Rightarrow \frac{1}{v} = - \frac{1}{60}$

or, v = − 60 cm
Hence, the required image will be located at a distance of 60 cm in front of the concave mirror.

Page No 412:

Answer:

Given,
Height of the object, h₁ = 20 cm,
Distance of image from screen v = −5.0 m = −500 cm,

Since, we know that
$- \frac{v}{u} = \frac{h_{2}}{h_{1}} or \frac{- (- 500)}{u} = \frac{50}{20}$
Where 'u' is the distance of object from screen.
(As the image is inverted)
$or u = - 500 \times \frac{2}{5} = - 200 cm = - 2.0 m$

Using mirror formula,
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} or \frac{1}{- 5} + \frac{1}{- 2} = \frac{1}{f}$
$or - \frac{1}{f} = \frac{7}{10} or f = - \frac{10}{7} = - 1.44 m$
Hence, the required focal length of the concave mirror is 1.44 m.

Page No 412:

Answer:

Using sign conventions, given,
Focal length of the concave mirror:
f = −20 cm
As per the question,
Magnification (m) is:
$m = - \frac{v}{u} = 2$
⇒ v = −2u

Case I (Virtual image):

Using mirror formula,
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{2 u} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{2 u} = \frac{1}{f} \Rightarrow u = \frac{f}{2} = 10 cm$

Case II (Real image)

Using mirror formula,
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow - \frac{1}{2 u} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{3}{2 u} = \frac{1}{f} \Rightarrow u = \frac{3 f}{2} = 30 cm$
Hence, the required positions of objects are 10 cm or 30 cm from the concave mirror.

Page No 412:

Answer:

Given,
Height of the object, h₁ = 1 cm
Focal length of the concave mirror, f = 7.5 cm = $\frac{15}{2} cm$
Magnification is given as,
$m = - \frac{v}{u} = \frac{h_{i}}{h_{0}} = 0.6 Using mirror equation, \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{0.6 u} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{5}{3 u} - \frac{1}{u} = \frac{2}{15} \Rightarrow u = 5 cm$

Hence, the distance of the object from the mirror is 5 cm.

Page No 412:

Answer:

Given,
Height (h₁) of the candle flame taken as object AB = 1.6 cm
Diameter of the ball bearing (d) = 0.4 cm
So, radius = 0.2 cm
Distance of object, u = 20 cm

Using mirror formula,

$\frac{1}{v} + \frac{1}{u} = \frac{2}{R}$

Putting the values according to sign conventions, we get,
$\frac{1}{(- 20)} + \frac{1}{v} = \frac{2}{0.2} \Rightarrow \frac{1}{v} = \frac{1}{20} + 10 \Rightarrow v = 0.1 cm or 1.0 mm insi de the ball bearing . Magnification = m = \frac{A' B'}{AB} = - \frac{v}{u} = m = \frac{A' B'}{200} = \frac{1}{200} \Rightarrow A' B' = \frac{AB}{200} = + \frac{1.6}{200} = + 0.08 cm (+ 0.008 cm)$

Hence, the distance of the image is 1 cm.
Height of the image is 0.008 cm.

Page No 412:

Answer:

Given,
Height of the object AB, h₁ = 3 cm
Distance of the object from the convex mirror, u = −7.5 cm
Focal length of the convex mirror (f) = 6 cm
Using mirror formula,
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{1}{u}$
Putting values according to sign convention,
$\frac{1}{v} = \frac{1}{6} - \frac{1}{(- 7.5)} \frac{1}{v} = \frac{1}{6} + \frac{1}{7.5} = \frac{3}{10} \Rightarrow v = \frac{10}{3} cm$
Magnification = m = $- \frac{v}{u} = \frac{10}{7.5 \times 3}$
$\Rightarrow \frac{A' B'}{AB} = \frac{10}{7.5 \times 3} \Rightarrow A' B' = \frac{100}{75} = \frac{4}{3} = 1.33 cm$
Where A'B' is the height of the image.

Hence, the required location of the image is $\frac{10}{3} cm$ from the pole and image height is 1.33 cm. Nature of the image is virtual and erect.

Page No 412:

Answer:

Given,
Radius of curvature of concave mirror, R = 20 cm
So its focal length will be f = $\frac{R}{2}$ = −10 cm
For part AB of the U shaped wire, PB = 30 + 10 = 40 cm

Therefore, u = −40 cm
Using mirror formula,
$\Rightarrow \frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} = - \frac{1}{10} - \frac{1}{- 40} = - \frac{3}{40}$
$\Rightarrow v = \frac{- 40}{3} = 13.3 cm$
$So, PB' = 13.3 cm$
$m = \frac{A' B'}{AB} = \frac{- (v)}{u}$
$= \frac{- (- 13.3)}{- 40} = - \frac{1}{3} \Rightarrow A' B' = - \frac{10}{3} = - 3.33 cm$

For part CD of the U shaped wire, PC = 30 cm
Therefore, u = −30 cm
Again, using mirror equation:
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$
$\frac{1}{v} = - \frac{1}{10} - \frac{1}{- 30} \frac{1}{v} = - \frac{1}{10} + \frac{1}{30} = - \frac{1}{15}$
⇒ v = −15 cm = PC'
$m = \frac{C' D'}{CD} = - \frac{v}{u}$
$= - \frac{(- 15)}{- 30} = - \frac{1}{2}$
⇒ C'D' = 5 cm
⇒ B'C' = PC' − PB'
= 15 − 13.3 = 1.7 cm

Hence, the total length of the U- shaped wire is A'B' + B'C' + C'D'
= (3.3) + (1.7) + 5 = 10 cm

Page No 412:

Answer:

Given,
Distance of the man's face (here, taken as object), u = −25 cm
According to the question, magnification, m = 1.4
$m = \frac{A' B'}{AB} = - \frac{v}{u} \Rightarrow 1.4 = - \frac{(v)}{- 25} \Rightarrow \frac{14}{10} = \frac{v}{25} \Rightarrow v = \frac{25 \times 14}{10} = 35 cm$

Using equation of mirror, we get:
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$\Rightarrow \frac{1}{f} = \frac{1}{35} - \frac{1}{25}$
$= \frac{5 - 7}{175} = - \frac{2}{175}$
⇒ f = −87.5

Hence, the required focal length of the concave mirror is 87.5 cm.

Page No 412:

Answer:

(a) reflection
(b) refraction

When the light strikes on a surface nearly parallel to it, it then bends by a small and fixed angle after reflection. Also, when the light travels from one medium to another with slight differences in their refractive indices, it bends by a small angle. Thus, the bending of light by a small but fixed angle can be the case of either reflection or refraction.

Page No 412:

Answer:

(b) If the final rays are converging, we have a real image.

This is because a real image is formed by converging reflected/refracted rays from a mirror/lens.

Page No 412:

Answer:

(a) Pole
(c) Radius of curvature
(d) Principal axis

Paraxial rays are the light rays close to the principal axis. The focus of the spherical mirror for paraxial rays are different from the focus for marginal rays. So, the focus depends on whether the rays are paraxial or marginal. The pole, radius of curvature and the principal axis of a spherical mirror do not depend on paraxial or marginal rays.

Page No 412:

Answer:

(c) the object is real but the image is virtual
(d) the object is virtual but the image is real

The virtual image of a real object and the real image of a virtual object are always erect.

Page No 412:

Answer:

(c) the image will not be shifted
(d) the intensity of the image will decrease

If the upper half portion of the convex lens is painted, then only the intensity of the image will decrease, as the amount of light passing through the lens will decrease. Also, there will be no shift in the position of the image because all the parameters remain the same.

Page No 412:

Answer:

(a) there will be an image at O₁
(b) there will be an image at O₂

The lens L₁ converges the light at point O₁ and lens L₂ converges the light at O₂. As the upper half of lens L₃ has a refractive index equal to that of L₁, it will converge the light at O₁ and thus the image will be formed at O₁. Also, the lower half of lens L₃ has a refractive index equal to that of lens L₂, it will converge the light at O₂ and thus the image will be formed at O₂.

Page No 412:

Answer:

(b) must be greater than 20 cm

Let the image be formed at a distance of x cm from the lens.
Therefore, the distance of the object from the lens, u, will be = (40 − x) cm
From lens formula:
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{f} = \frac{1}{x} - \frac{1}{40 - x} \Rightarrow \frac{1}{f} = \frac{40 - x - x}{40 x - x^{2}} \Rightarrow f = \frac{40 x - x^{2}}{40 - 2 x} \Rightarrow f (40 - 2 x) = 40 x - x^{2} \Rightarrow x^{2} - 2 f x - 40 x + 40 f = 0 \Rightarrow x^{2} - (2 f + 40) x + 40 f = 0$

Therefore, we get x as:

$x = \frac{(2 f + 40) \pm \sqrt{(2 f + 40)^{2} - 160 f}}{2}$

Page No 413:

Answer:

Given,
Focal length of the concave mirror, f = − 7.6 m
Distance between earth and moon taken as object distance, u = −3.8 × 10⁵ km
Diameter of moon = 3450 km
Using mirror equation:
$\Rightarrow \frac{1}{v} = \frac{1}{u} + \frac{1}{f}$
$∴ \frac{1}{v} + (- \frac{1}{3.8 \times 10^{8}}) = (- \frac{1}{7.6})$

As we know, the distance of moon from earth is very large as compared to focal length it can be taken as ∞.
Therefore, image of the moon will be formed at focus, which is inverted.
$\Rightarrow \frac{1}{v} = - \frac{1}{7.6}$
⇒ v = −7.6 m

We know that magnification is given by:
$m = - \frac{v}{u} = \frac{d_{image}}{d_{object}}$
$\Rightarrow \frac{- (- 7.6)}{(- 3.8 \times 10^{8})} = \frac{d_{i m a g e}}{3450 \times 10^{3}}$
$d_{image} = - \frac{3450 \times 7.6 \times 10^{3}}{3.8 \times 10^{8}}$
= $-$ 0.069 m = $-$ 6.9 cm
Hence, the required diameter of the image of moon is 6.9 cm.

Page No 413:

Answer:

Given,
Distance of the circle from the mirror taken as object distance, u = −30 cm,
Focal length of the concave mirror, f = −20 cm

Using mirror formula,
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$\Rightarrow \frac{1}{v} + (- \frac{1}{30}) = - \frac{1}{20}$

$\Rightarrow \frac{1}{v} = \frac{1}{30} - \frac{1}{20} = \frac{1}{60}$
⇒ v = $-$ 60 cm
Therefore, image of the circle is formed at a distance of 60 cm in front of the mirror.
We know magnification (m) is given by:
$m = - \frac{v}{u} = \frac{R_{image}}{R_{object}}$
$\Rightarrow - \frac{(- 60)}{(- 30)} = \frac{R_{image}}{2}$
Where R_object and R_image are radius of the object and radius of the image, respectively.
⇒ R_image = 4 cm
Hence, the required radius of the circle formed by the image is 4 cm.

Page No 413:

Answer:

Given,
A concave mirror of radius 'R' kept on a horizontal table.
'h' is height up to which the water is poured into the concave mirror.
Let the object be placed at height 'x' above the surface of water.
We know if we place the object at the centre of curvature of the mirror, then the image itself will be formed at the centre of curvature.
Therefore, the apparent position of the object with respect to the mirror should be at the centre of curvature so that the image is formed at the same position.

Since,
$\frac{Real depth}{Apparent depth} = \frac{1}{μ}$
(with respect to mirror)
$Now, \frac{x}{R - h} = \frac{1}{μ} \Rightarrow x = \frac{R - h}{μ}$
Hence, the object should be placed at $\frac{R - h}{μ}$ above the water surface.

Page No 413:

Answer:

Given,
Two converging mirrors having equal focal length 'f '.
Both the mirrors will produce one image under two conditions:

Case-1
When the point source is at the centre of curvature of the mirrors, i.e, at a distance of '2f ' from each mirror, the images will be produced at the same point 'S'. Therefore, d = 2f + 2f = 4f

Case-II
When the point source 'S' is at focus, i.e., at a distance 'f ' from each mirror, the rays from the source after reflecting from one mirror will become parallel and so these parallel rays, after the reflection from the other mirror the object itself. Thus, only one image is formed.
Here d = f + f = 2f

Page No 413:

Answer:

Given,
Converging mirror M₁ with focal length (f₁) = 20 cm
Converging mirror M₂ with focal length (f₂) = 20 cm
f₁ = f₂ = 20 cm = f
Point source is at a distance of 30 cm from M₁.

As the 1^st reflection is through the mirror M₁,
u = −30 cm
f = −20 cm

Using mirror equation:
$\Rightarrow \frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$\Rightarrow \frac{1}{v} + \frac{1}{- 30} = - \frac{1}{20}$
$\Rightarrow \frac{1}{v} = - \frac{1}{20} + \frac{1}{30} \Rightarrow \frac{1}{v} = - \frac{1}{60} \Rightarrow v = - 60 cm$

and for the 2^nd reflection at mirror M₂,
u = 60 − (30 + x) = 30 − x
v = − x, f = 20 cm

Again using the mirror equation:
$\Rightarrow \frac{1}{30 - x} - \frac{1}{x} = \frac{1}{20}$

$\Rightarrow \frac{x - 30 + x}{x (30 - x)} = \frac{1}{20}$
⇒ 40x − 600 = 30x − x²
⇒ x² + 10x − 600 = 0

$\Rightarrow x = \frac{10 + 50}{2} = \frac{40}{2}$
= 20 cm or − 30 cm
∴ Total distance between the two lines is 20 + 30 = 50 cm

(b) Location of the image formed by the single reflection from M₂ = 60 $-$ 50 = 10

Thus, the image formed by the single reflection from M₂ is at a distance of 10 cm from mirror M₂_.

Page No 413:

Answer:

Given,
Angle of incidence, i = 45°
Angle of refraction, r = 30°
Using Snell's law,
$\frac{\sin i}{\sin r} = \frac{3 \times 10^{8}}{v}$
$= \frac{\sin 45 °}{\sin 30 °} = \frac{(\frac{1}{\sqrt{2}})}{(\frac{1}{2})}$
$= \frac{2}{\sqrt{2}} = \sqrt{2}$
$So, v = \frac{3 \times 10^{8}}{\sqrt{2}} m / s$

Let x be the distance travelled by light in the slab.
Now,
$x = \frac{1 m}{\cos 30 °} = \frac{2}{\sqrt{3}} m$
We know:
Time taken $= \frac{Distance}{Speed}$
$= \frac{2}{\sqrt{3}} \times \frac{\sqrt{2}}{3 \times 10^{8}}$
= 0.54 × 10⁻⁸
= 5.4 × 10⁻⁹ s

Page No 413:

Answer:

Given,
Length of the pole = 1.00 m
Water level of the swimming pool is 50.0 cm higher than the bed.
Refractive index (μ) of water = 1.33

According to the figure, shadow length = BA' = BD + DA'= 0.5 + 0.5 tan r
Using Snell's law:
$Now 1.33 = \frac{\sin 45 °}{\sin r} \Rightarrow \sin r = \frac{1}{1.33 \sqrt{2}} = 0.53 \Rightarrow \cos r = \sqrt{1 - \sin^{2} r} = \sqrt{1 - (0.53)^{2}} = 0.85 So, \tan r = 0.6235$

Therefore, shadow length of the pole = (0.5) $\times$ (1 + 0.6235) = 0.81175 m
= 81.2 cm

Page No 413:

Answer:

Given,
Depth of the lake = 2.5 m
Refractive index (μ) of water = $\frac{4}{3}$
When the sun is just setting, θ is approximately = 90˚

Therefore, incidence angle is 90˚
Using Snell's law:

$∴ \frac{\sin i}{\sin r} = \frac{μ_{2}}{μ_{1}}$
$\Rightarrow \frac{1}{\sin r} = \frac{\frac{4}{3}}{1}$
$\Rightarrow \sin r = \frac{3}{4}$
$\Rightarrow r = 49 °$

From the figure, W'O = x is the distance of the shadow
$Thus, (\frac{x}{2.5}) = \tan r = 1.15$
$\Rightarrow x = 2.5 \times 1.15 = 2.8 m$

Page No 413:

Answer:

Given,
Thickness of the glass slab, d = 2.1 cm
Refractive index, μ = 1.5
Shift due to the glass slab is given by,
$∆ t = [1 - (\frac{1}{μ})] d$
$= [1 - (\frac{1}{1.5})] (2.1)$
$= \frac{1}{3} (2.1) = 0.7 cm$
Hence, the microscope should be shifted by 0.70 cm to focus the object 'P' again.

Page No 413:

Answer:

Given,
Height of water, d_w = 20 cm
Height of oil, d_O = 20 cm
The refractive index of the water (μ_w) = 1.33
The refractive index of oil (μ_O) = 1.30

Shift due to water is given by,
$∆ t_{w} = 1 - (\frac{1}{μ_{w}}) d_{w}$
$= [1 - (\frac{1}{1.33})] 20$
$= \frac{1}{4} (20) = 5 cm$

Shift due to oil,
$∆ t_{O} = [1 - (\frac{1}{1.3})] 20 = 4.6 cm$

Therefore, total shift, Δt = 5 + 4.6 = 9.6 cm
Hence, apparent depth = 40 − (9.6) = 30.4 cm between the surface.

Page No 413:

Answer:

Given,
From the figure we can infer that the air is present in between the sheet, so it does not affect the shift. Therefore, the shift is only due to 3 sheets of different refractive indices, which is given by:

$∆ t = [1 - \frac{1}{μ_{1}}] t_{1} + [1 - \frac{1}{μ_{2}}] t_{2} + [1 - \frac{1}{μ_{3}}] t_{3}$
$= [1 - (\frac{1}{12})] (0.2) + [1 - (\frac{1}{1.3})] (0.3) + [1 - (\frac{1}{1.4})] (0.4)$
= 0.2 cm
Hence, location of image of point P is located 0.2 cm above point P.

Page No 413:

Answer:

k number of transparent slabs are arranged one over the other.
Refractive indices of the slabs = μ₁, μ₂, μ₃, ..., μ_k
Thickness of the slabs = t₁, t₂, t₃,.., t_k
Shift due to one slab:
$∆ t = [1 - \frac{1}{μ}] t$

For the combination of multiple slabs, the shift is given by,
$∆ t = [1 - \frac{1}{μ_{1}}] t_{1} + [1 - \frac{1}{μ_{2}}] t_{2} + . . . + [1 - \frac{1}{μ_{k}}] t_{k} . . . (i)$

Let μ be the refractive index of the combination of slabs.
The image is formed at the same place.
So, the shift will be:

$Δ t = [1 - (\frac{1}{μ})] (t_{1} + t_{2} . . . + t_{k}) . . . (ii)$

Equating (i) and (ii), we get:

$[1 - (\frac{1}{μ})] (t_{1} + t_{2} . . . + t_{k}) = [1 - \frac{1}{μ_{1}}] t_{1} + [1 - \frac{1}{μ_{2}}] t_{2} + [1 - \frac{1}{μ_{k}}] t_{k} = (t_{1} + t_{2} . . . + t_{k}) - (\frac{t_{1}}{μ_{1}} + \frac{t_{2}}{μ_{2}} + \frac{t_{k}}{μ_{k}}) = - \frac{1}{μ} \sum_{i = 1}^{k} t_{i} = - \sum_{i = 1}^{k} (\frac{t_{i}}{μ_{i}}) \Rightarrow μ = \frac{\sum_{i = 1}^{k} t_{i}}{- \sum_{i = 1}^{k} (\frac{t_{i}}{μ_{i}})}$
Hence, the required equivalent refractive index is $\frac{\sum_{i = 1}^{k} t_{i}}{\sum_{i = 1}^{k} (\frac{t_{i}}{μ_{i}})}$ .

Page No 413:

Answer:

Given,
Diameter of the cylindrical vessel = 12 cm
∴ radius r = 6 cm
Diameter of the cylindrical glass piece = 8 cm
∴ radius r^_'= 4 cm and its height, h₁ = 8 cm
Refractive index of glass, μ_g = 1.5
Refractive index of water, μ_w = 1.3
Let h be the final height of the water column.

The volume of the cylindrical water column after the glass piece is put will be:
$π$ r²h = 800π + $π$ r'²h₁
or r²h = 800 + r_'²h₁
or (6)² h = 800 + (4)² 8
$or h = \frac{800 + 128}{36} = \frac{928}{36} = 25.7 cm$

There will be two shifts; due to the glass block as well as water:

$∆ t_{1} = [1 - \frac{1}{μ_{g}}] t_{g} = [1 - \frac{1}{\frac{3}{2}}] \times 8 = 2.26 cm ∆ t_{2} = [1 - \frac{1}{μ_{w}}] t_{w} = [1 - \frac{1}{\frac{4}{3}}] \times (25.7 - 8) = 4.44 cm$

Hence, the total shift = (Δt₁+ Δt₂) = (2.66 + 4.44) cm = 7.1 cm above the bottom.

Page No 414:

Answer:

Given,
Refractive index of water = μ.
Height of the pot = H
Let us take x as the distance of the image of the eye formed above the surface of the water as seen by the fish.

We can infer from from the diagram,

$\frac{H}{x} = \frac{Real depth}{Apparent depth} = \frac{1}{μ} or x = μ H$
The distance of the direct image $= \frac{H}{2} + μ H = H (μ + \frac{1}{2})$
Similarly, image through mirror = $\frac{H}{2} + (H + x) = \frac{3 H}{2} + μ H = H (\frac{3}{2} + μ)$
b) We know that:
$\frac{\frac{H}{2}}{y} = μ, S o y = \frac{H}{2 μ}$
Where, y is the distance of the image of the fish below the surface as seen by the eye.
Direct image = $H + y = H + \frac{H}{2 μ} = H (1 + \frac{1}{2 μ})$
Again another image of fish will be formed H/2 below the mirror.
Real depth for that image of fish becomes H + H/2 = 3H/2
So, apparent depth from the surface of water = 3H/2μ

So, distance of the image from the eye $= \frac{H}{2} + \frac{3 H}{2 μ} = H (1 + \frac{3}{2 μ})$

Page No 414:

Answer:

Given,
Refractive index of water μ = 1.33
Radius of the cylindrical vessel = 3 cm
Height of the cylindrical vessel = 4 cm

Let x be the length of BD
According to the diagram,
$\frac{x}{3} = c o t r . . . . (i)$

Using Snell's law,
$\frac{\sin i}{\sin r} = \frac{1}{1.33} = \frac{3}{4} \Rightarrow \sin r = \frac{4}{3} \sin i = \frac{4}{3} \times \frac{3}{5} = \frac{4}{5} As \sin i = \frac{B C}{A C} = \frac{3}{5} \Rightarrow \cot r = \frac{3}{4} . . . . . (ii)$
From (i) and (ii),

$\frac{x}{3} = \frac{3}{4} \Rightarrow x = \frac{9}{4} = 2.25 cm$

Hence, the ratio of real and apparent depth of the image will be $4 : 2.25 = 1.78$

Page No 414:

Answer:

Given,
Diameter and height (h) of the cylindrical vessel = 30 cm
Therefore, its radius (r) = 15 cm
We know the refractive index of water (μ_w) = 1.33 $= \frac{4}{3}$

Using Snell's law,

$\frac{\sin i}{\sin r} = \frac{1}{μ_{w}} \frac{\sin i}{\sin r} = \frac{3}{4} As r = 45 ° \Rightarrow \sin i = \frac{3}{4 \sqrt{2}}$

Point P will be visible when the refracted ray makes an angle of 45˚ at the point of refraction.
Let x be the distance of point P from X.
$\tan 45 ° = \frac{x + 10}{d} \Rightarrow d = x + 10 . . . (i) and \tan i = \frac{x}{d} \Rightarrow \frac{3}{\sqrt{23}} = \frac{d - 10}{d} (∵ \sin i = \frac{3}{4 \sqrt{2}} \Rightarrow \tan i = \frac{3}{\sqrt{23}}) \Rightarrow \frac{3}{\sqrt{23}} - 1 = \frac{- 10}{d} \Rightarrow d = \frac{\sqrt{23} \times 10}{\sqrt{23} - 3} d = 26.7 cm$

Hence, the required minimum height of water = 26.7 cm

Page No 414:

Answer:

Given,
Angle of incidence = 45˚
Thickness of the plate = $\sqrt{2} cm$
Refractive index (μ) of the plate = 2.0

Applying Snell's law,

$\frac{\sin i}{\sin r} = \frac{1}{μ} \Rightarrow \frac{\sin i}{\sin r} = \frac{2}{1} \Rightarrow \sin r = \frac{\sin 45 °}{2} \Rightarrow \sin r = \frac{1}{2 \sqrt{2}} \Rightarrow r = 21 °$

Therefore, $θ = (45 ° - 21 °) = 24 °$

From figure, we know that BD is the shift in path which is equal to (AB)sin 24˚
$= 0.406 \times AB = \frac{AE}{\cos 21 °} \times 0.406 = 0.62 cm$

Hence, the required shift in the path of light is 0.62 cm.

Page No 414:

Answer:

Given,
Refractive index of the optical fibre is represented by μ_o = 1.72
Refractive index of glass coating is represented by μ_g= 1.50
Let the critical angle for glass be θ_c

Using the Snell's law,

$\frac{\sin i}{\sin r} = \frac{\sin θ_{c}}{\sin 90 °} = \frac{μ_{g}}{μ_{0}} \Rightarrow \frac{\sin θ_{c}}{\sin 90 °} = \frac{1.50}{1.72} = \frac{75}{86} \Rightarrow θ_{c} = \sin^{- 1} (\frac{75}{86})$

Hence, the required critical angle is $\sin^{- 1} (\frac{75}{86})$

Page No 414:

Answer:

Given,
Refractive index (μ) of prism = 1.50
Let us take θ_c as the critical angle for the glass.

So, According to Snell's law,
$\frac{\sin θ_{c}}{\sin 90 °} = \frac{1}{μ}$
$\Rightarrow \sin θ c = \frac{1}{1.5} = \frac{2}{3}$
$\Rightarrow θ_{c} = \sin^{- 1} \frac{2}{3}$

Condition for total internal reflection: 90° − $ϕ$ > θ_c
⇒ $ϕ$ < 90° $-$ θ_c
$\Rightarrow ϕ < 90 ° - \sin^{- 1} (\frac{2}{3})$
$\Rightarrow ϕ < \cos^{- 1} (\frac{2}{3})$
Hence, the largest angle for which light is totally reflected at the surface AC is $\cos^{- 1} (\frac{2}{3})$ .

Page No 414:

Answer:

Given,
Refractive index (μ) of the glass = 1.5
We know from the definition of a critical angle (θ_c) that if a refracted angle measures more than 90°, then total internal reflection takes place.
Hence, the maximum angle of refraction is 90°.

Page No 414:

Answer:

Given,
Refractive index of glass, μ_g = 1.5
Refractive index of air, μ_a= 1.0
Angle of incidence 0° < i < 90
Let us take θ_c as the Critical angle
$\Rightarrow \frac{\sin θ c}{\sin r} = \frac{μ_{a}}{μ_{g}}$
$\Rightarrow \frac{\sin θ_{c}}{\sin 90 °} = \frac{1}{1.5} = 0.66 \Rightarrow \sin θ_{c} = 0.66 \Rightarrow θ_{c} = \sin^{- 1} (0.66)$
⇒ θ_c = 40°48"

The angle of deviation (δ) due to refraction from glass to air increases as the angle of incidence increases from 0° to 40°48". The angle of deviation (δ) due to total internal reflection further increases from 40°48" to 45° and then it decreases, as shown in the graph.

Page No 414:

Answer:

Given,
Refractive index of glass $: μ_{g} = 1.5 = \frac{3}{2}$
Refractive index of water $: μ_{w} = 1.33 = \frac{4}{3}$
As per the question,
For two angles of incidence,

1. When light passes straight through the Normal,
⇒ Angle of incidence = 0°
⇒ Angle of refraction = 0°
⇒ Angle of deviation = 0°

2. When light is incident at critical angle θ_c,
$\frac{\sin θ_{c}}{\sin r} = \frac{μ_{w}}{μ_{g}}$
(since the light is passing from glass to water)
$\Rightarrow \sin θ_{c} = \frac{8}{9}$
$\Rightarrow θ_{c} = \sin^{- 1} (\frac{8}{9}) = 62.73 °$
⇒ Angle of deviation
=90° − θ_c
$= 90 - \sin^{- 1} \frac{8}{9}$
$= \cos^{- 1} \frac{8}{9}$
=37.27°

Here, if the angle of incidence increased beyond the critical angle, total internal reflection occurs and deviation decreases.
Therefore, the range of angle of deviation is in between 0 to 37.27° or $\cos^{- 1} (\frac{8}{9})$ .

Page No 414:

Answer:

Given,
Light falls from glass to air.
Refractive index (μ) of glass = 1.5

Critical angle (θ_c) $= \sin^{- 1} (\frac{1}{μ})$
$= \sin^{- 1} (\frac{1}{1.5}) = 41.80$

We know that the maximum attainable angle of deviation in refraction is (90° − 41.8°)
= 47.2°
In this case, total internal reflection must have taken place.
In reflection,
Deviation = 180° − 2i = 90°
⇒ 2i = 90°
⇒ i = 45°
Hence, the required angle of incidence is 45°.

Page No 414:

Answer:

Given,
Refractive index is μ

(a)
Let the point source be P, which is placed at a depth of h from the surface of water.
Let us take x as the radius of the circular area.
and let θ_c be the critical angle.

Thus,
$\frac{x}{h} = \tan θ_{c} \frac{x}{h} = \frac{\sin θ_{c}}{\sqrt{1 - \sin^{2} θ_{c}}} = \frac{\frac{1}{μ}}{\sqrt{1 - \frac{1}{μ^{2}}}} (∵ \sin θ_{c} = \frac{1}{μ}) \frac{x}{h} = \frac{1}{\sqrt{μ^{2} - 1}} x = \frac{h}{\sqrt{μ^{2} - 1}}$

Clearly from figure, the light escapes through a circular area at a fixed distance r on the water surface, directly above the point source.
That makes a circle, the centre of which is just above P.

(b)
The angle subtended by the radius of the circular area on the point source P:
$\Rightarrow \sin θ_{c} = \frac{1}{μ}$
$\Rightarrow θ_{c} = \sin^{- 1} (\frac{1}{μ})$

Page No 414:

Answer:

Given,
Height (h) of the water in the container = 20 cm
Ceiling of the room is 2.0 m above the water surface.
Radius of the rubber ring = r
Refractive index of water = 4/3

(a)

From the figure, we can infer:
$\sin i = \frac{15}{25}$

Using Snell's law, we get:

$\frac{\sin i}{\sin r} = \frac{1}{μ} = \frac{3}{4} \Rightarrow \sin i = \frac{4}{5}$

From the figure, we have:

$\tan r = \frac{x}{2} So, \sin r = \frac{\tan r}{\sqrt{1 + \tan^{2} r}} = \frac{\frac{x}{2}}{\sqrt{1 + \frac{x^{2}}{4}}} \Rightarrow \frac{x}{\sqrt{4 + x^{2}}} = \frac{4}{5}$
$\Rightarrow 25 x^{2} = 16 (4 + x^{2}) \Rightarrow 9 x^{2} = 64 \Rightarrow x = \frac{8}{3} m$
Total radius of the shadow = $\frac{8}{3} + 0.15 = 2.81 m$

(b)
Condition for the maximum value of r:
Angle of incidence should be equal to the critical angle, i.e., $i = θ_{c}$ .
Let us take R as the maximum radius.
Now,
$\sin θ_{c} = \frac{\sin θ_{c}}{\sin r} = \frac{R}{\sqrt{R^{2} + 20}} = \frac{3}{4} (\sin r = 1) \Rightarrow 16 R^{2} = 9 R^{2} + 9 \times 400 \Rightarrow 7 R^{2} = 9 R^{2} + 9 \times 400 \Rightarrow R = 22.67 cm$

Page No 414:

Answer:

Given,
Refractive index (μ) of the material from which prism is made = 1.732
We know refractive index is given by:

$μ = \frac{\sin [\frac{δ_{\min} + A}{2}]}{\sin [\frac{A}{2}]}$

Where δ_min is the angle of minimum deviation and A is the angle of prism = 60˚
$\Rightarrow 1.732 \times \sin (30 °) = \sin (\frac{δ_{\min} + 60 °}{2})$

$\Rightarrow \frac{1.732}{2} = \sin (\frac{δ_{\min} + 60 °}{2})$
$\Rightarrow (\frac{δ_{\min} + 60 °}{2}) = 60 °$
δ_min = 60°
δ_min = 2i − A
2i = 120°
i = 60°

Hence, the required angle of deviation is 60°.

Page No 414:

Answer:

Given,
The refractive index of the prism material (μ) = 1.5
Angle of prism form the figure = 4˚

We know that,
$μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}} μ = \frac{(\frac{A + δ_{m}}{2})}{\frac{A}{2}} (∵ f o r s m a l l a n g l e \sin θ \approx θ) = \frac{A + δ_{m}}{2 °} 1.5 = \frac{4 ° + δ_{m}}{2 °} δ_{m} = 2 °$

Hence, the angle of deviation is 2˚

Page No 414:

Answer:

Given,
The angle of the prism (A) = 60˚
The angle of deviation (δ_m) = 30˚
$Refractive index, μ \leq \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})} μ \leq \frac{\sin (\frac{60 ° + δ_{m}}{2})}{\sin (\frac{60 °}{2})} μ \leq 2 \sin (\frac{60 ° + δ_{m}}{2})$
As there is one ray that has been found which has deviated by 30˚, the angle of minimum deviation should be either equal to or less than 30˚ but it can not be more than 30˚.

Therefore,
$μ \leq 2 \sin (\frac{60 ° + δ_{m}}{2}) μ \leq 2 \sin (\frac{60 ° + 30 °}{2}) = 2 \sin (45 °)$

Refractive index (μ) will be more if angle of deviation (δ_m) is more.
$μ \leq 2 \times \frac{1}{\sqrt{2}}$
$\Rightarrow μ \leq \sqrt{2}$

Hence, the required limit of refractive index is $\sqrt{2}$

Page No 414:

Answer:

Given,
Let the refractive indices of two mediums be μ₁=1.0 and μ₂ =1.5
Point C is the centre of curvature, the distance between C and the pole is 20 cm.
Therefore, radius of curvature (R) = 20 cm
Distance between source S and pole is 25 cm.
Therefore, object distance (u) = −25

Using lens equation,

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$

$\Rightarrow \frac{1.5}{v} - \frac{1}{(- 25)} = \frac{0.5}{20}$

$\Rightarrow \frac{1.5}{v} = \frac{- 3}{200}$

$∴ v = - \frac{200 \times 1.5}{0.3 \times 10} = - 100$

Hence, the required location of the image is 100 cm from the pole and on the side of S.

Page No 414:

Answer:

Given,
Spherical surface of radius (R) = 30 cm
Medium A has refractive index (μ₁) = 1.33
Medium B has refractive index (μ₂) = 1.48
Medium A is the convex side of surface.

Since,
We know that paraxial rays become parallel after refraction
i.e, the image of the point object will be formed at infinity.
Therefore v = ∞

Using the lens equation,
$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R} \Rightarrow \frac{1.48}{\infty} - \frac{1.33}{u} = \frac{1.48 - 1.33}{30} \Rightarrow - \frac{1.33}{u} - \frac{0.15}{30} ∴ u = - 266.0 cm$

Hence, the object is placed at a distance of 266.0 cm from the convex surface on side A.

Page No 415:

Answer:

Given,
The radius of the transparent hemisphere (R) = 3.0 cm
Refractive index of the material (μ₂) = 2.0
Let the critical angle be θ_c
∴ critical angle is given by θ_c = $\sin^{- 1} (\frac{1}{μ_{2}}) = \sin^{- 1} (\frac{1}{2}) = 30 °$
(a)
From the figure it is seen that the angle of incidence is greater than the critical angle, so the rays are totally reflected at the plane surface.

(b)
Using the lens equation:

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R} \Rightarrow \frac{2}{v} - (- \frac{1}{\infty}) = \frac{2 - 1}{3} (for parallel rays u = \infty) \Rightarrow \frac{2}{v} = \frac{1}{3} \Rightarrow v = 6 cm$

If we complete the sphere then the image will be formed diametrically opposite to A.

(c)
By internal reflection, the image is formed in front of A.

Page No 415:

Answer:

Given,
Radius of the sphere = 5.0 cm
Refractive index of the sphere (μ₁) = 1.5
An object is embedded in the glass sphere 1.5 cm left to the centre.

(a)
When the image is seen by observer from left of the sphere,
from surface the object distance (u) = − 3.5 cm
μ₁ = 1.5
μ₂= 1
v₁ = ?

Using lens equation:

$\frac{μ_{2}}{v_{1}} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$
$\frac{1}{v_{1}} - \frac{1.5}{- (3.5)} = \frac{- 0.5}{- 5}$
$\frac{1}{v_{1}} = \frac{1}{10} - \frac{3}{7}$
$= \frac{7 - 30}{70}$
$= \frac{- 23}{70}$
$v_{1} = \frac{- 70}{23} ≃ - 3 cm$
So the image will be formed at 2 cm (5 cm $-$ 3cm) left to centre.

(b)

When the image is seen by observer from the right of the sphere,
u = −(5.0 + 1.5) = − 6.5,
R = −5.00 cm
μ₁ = 1.5, μ₂ = 1, v = ?

Using lens equation:

$\frac{μ_{2}}{v_{1}} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$
$\frac{1}{v} - \frac{1.5}{6.3} = \frac{1}{10}$
$\frac{1}{v} = \frac{1}{10} - \frac{3}{13}$
$= \frac{13 - 30}{130}$

$= \frac{- 17}{130}$

$v = \frac{- 130}{17} = - 7.65 cm$

Therefore, the image will be formed 7.6 − 5 = 2.6 towards left from centre.

Page No 415:

Answer:

Given,
Biconvex lens with each surface has a radius (R₁= R₂ = R) = 10 cm,
and thickness of the lens (t) = 5 cm
Refractive index of the lens (μ) = 1.50
Object is at infinity (∴ u = ∞ )

First refraction takes place at A.
We know that,

$\frac{μ_{g}}{v} - \frac{μ_{a}}{u} = \frac{μ_{g} - μ_{a}}{R} \Rightarrow \frac{1.5}{v} - (- \frac{1}{\infty}) = \frac{1.5 - 1}{10} \Rightarrow \frac{1.5}{v} = \frac{0.5}{10} \Rightarrow v = 30 cm$

Now, the second refraction is at B.
For this, a virtual object is the image of the previous refraction.
Thus, u = (30 − 5) = 25 cm

$\frac{μ_{a}}{v} - \frac{μ_{g}}{u} = \frac{μ_{a} - μ_{g}}{R} \Rightarrow \frac{1}{v} - \frac{1.5}{25} = \frac{1 - 1.5}{10} \Rightarrow \frac{1}{v} - \frac{1.5}{25} = \frac{- 0.5}{10} \Rightarrow v = 9.1 cm$

Hence, the image is formed 9.1 cm far from the second refraction or second surface of the lens.

Page No 415:

Answer:

Given,
The radius of the transparent sphere = r
Refraction at convex surface.
As per the question,
u = −∞, μ₁ = 1, μ₂ = ?

(a)
When image is to be focused on the surface,
Image distance (v) = 2r, Radius of curvature (R) = r
We know that,
$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R} \Rightarrow \frac{μ_{2}}{2 r} - (\frac{1}{- \infty}) = \frac{μ_{2} - 1}{r} \Rightarrow \frac{μ_{2}}{2 r} = \frac{μ_{2} - 1}{r} \Rightarrow μ_{2} = 2 μ_{2} - 2 \Rightarrow μ_{2} = 2$

(b)
When the image is to be focused at the centre,
Image distance (v) = r, Radius of curvature (R) = r
$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R} \Rightarrow \frac{μ_{2}}{r} - (\frac{1}{- \infty}) = \frac{μ_{2} - 1}{r} \Rightarrow \frac{μ_{2}}{r} = \frac{μ_{2} - 1}{r} \Rightarrow μ_{2} = μ_{2} - 1$

The above equation is impossible.
Hence, the image cannot be focused at centre.

Page No 415:

Answer:

Given,
Radius (R) of the cylindrical rod = 1.0 cm
Refractive index (μ_g) of the rod = 1.5 = $\frac{3}{2}$
Refractive index (μ_w) of water = 4/3

We know:

$\frac{μ_{g}}{v} - \frac{μ_{w}}{u} = \frac{μ_{g} - μ_{w}}{R}$
As per the question, u = −8 cm.
Now,
$\frac{3}{2 v} - (- \frac{4}{3 \times 8}) = \frac{\frac{3}{2} - \frac{4}{3}}{1} \Rightarrow \frac{3}{2 v} + \frac{1}{6} = \frac{1}{6} \Rightarrow v = \infty$

Hence, the image will be formed at infinity (∞).

Page No 415:

Answer:

Given:
Radius of the paperweight (R) = 3 cm
Refractive index of the paperweight (μ₂) = 3/2
Refractive index of the air (μ₁) = 1

In the first case, the refraction is at A.
u = 0 and R = ∞
We know:

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R} \Rightarrow \frac{\frac{3}{2}}{v} - \frac{1}{0} = \frac{μ_{2} - μ_{1}}{\infty} \Rightarrow v = 0$

Therefore, the image of the letter is formed at the point.
For the second case, refraction is at point B.
Here,
Object distance, u = −3 cm
R = − 3 cm
μ₁ = 3/2
μ₂ = 1
Thus, we have:

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R} \frac{1}{v} - \frac{3}{2 \times (- 3)} = \frac{1 - \frac{3}{2}}{- 3} \Rightarrow \frac{1}{v} + \frac{3}{2 \times 3} = \frac{1}{6} \Rightarrow \frac{1}{v} = \frac{1}{6} - \frac{1}{2} \Rightarrow v = - 3 cm$

Hence, there will be no shift in the final image.

Page No 415:

Answer:

Given,
Taking the radius of the paperweight as its thickness = 3 cm
Refractive index of the paperweight (μ_g) = 3/2
Refractive index of the air (μ₁) = 1

Image shift is given by:
$∆ t = (1 - \frac{1}{μ}) t = (1 - \frac{2}{3}) 3 = (\frac{1}{3}) 3 = 1 cm$

The upper surface of the paperweight is flat and the spherical spherical surface is in contact with the printed letter.
Therefore, we will take it as a simple refraction problem.

Hence, the image will appear 1 cm above point A.

Page No 415:

Answer:

As shown in the figure, OQ = 3r and OP = r
Thus, PQ = 2r
For refraction at APB,
we know that,
$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R} \Rightarrow \frac{1.5}{v} - (\frac{1}{- 2 r}) = \frac{0.5}{r} = \frac{1}{2 r} (∵ u = - 2 r) \Rightarrow v = \infty$

For the reflection in the concave mirror,
u = ∞
Thus, v = focal length of mirror = r/2
For the refraction of APB of the reflected image,
u = −3r/2
$\Rightarrow \frac{1}{v} - \frac{1.5}{3 r / 2} = \frac{- 0.5}{- r} (Here, μ_{1} = 1.5, μ_{2} = 1 a n d R = - r) \Rightarrow v = - 2 r$

As negative sign indicates images formed inside APB, so image should be at C.
Therefore, the final image is formed on the reflecting surface of the sphere.

Page No 415:

Answer:

(a) Let F be the the focal length of the given concavo-convex lens. Then,

$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{m}} + \frac{1}{f_{1}}$
   $= \frac{2}{f_{1}} + \frac{1}{f_{m}}$               $[∵ \frac{1}{f_{m}} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})]$

$\frac{1}{F} = \frac{2}{30} + \frac{1}{15} = \frac{2}{15}$
⇒ F = 7.5 cm

Hence, R = 15 cm

Therefore, the pin should be placed at a distance of 15 cm from the lens.

(b) If the concave part is filled with water,
      For focal length F'
       $\frac{1}{F'} = \frac{2}{f_{w}} + \frac{2}{f_{1}} + \frac{1}{f_{m}}$
    $= \frac{2}{90} + \frac{2}{30} + \frac{1}{15}$        $[∵ \frac{1}{f_{w}} = (\frac{4}{3} - 1) (+ \frac{1}{30})]$
$∴ F' = \frac{45}{7} cm$
Thus, pin should be placed at a distance of $\frac{90}{7}$ cm from the lens.

Page No 415:

Answer:

Given,
Double convex lens of focal length, f = 25 cm
Refractive index of the material, μ = 1.5
As per the question, the radius of curvature of one surface is twice that of the other.
i.e. R₁= R and R₂= −2R
(according to sign conventions)

Using the lens maker formula $\frac{1}{f} = (μ - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$ , we have:

$\frac{1}{25} = (1.5 - 1) [\frac{1}{R} - (\frac{1}{- 2 R})]$
$\Rightarrow \frac{1}{25} = 0.5 [\frac{3 R}{2}] \Rightarrow \frac{1}{25} = [\frac{3 R}{4}] \Rightarrow R = 18.75 cm$

R₁ = 18.75 cm and R₂ = 37.5 cm

Hence, the required radii of the double convex lens are 18.75 cm and 37.5 cm.

Page No 415:

Answer:

Given,
Radii of curvature of a lens, (R₁₎ = +20 cm and R₂ = +30 cm
Refractive index of the material of the lens, (μ) = 1.6
Refractive index of water, (μ_water) = 1.33

(a)
When the lens is placed in air,
Using lens maker formula:

$\frac{1}{f} = (μ - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$
$\frac{1}{f} = 0.6 [\frac{1}{20} - \frac{1}{30}]$
$f = \frac{0.6}{1060 \times 10}$
f = 100 cm
Thus, the focal length of the lens is 100 cm when it is placed in air.

(b)When the lens is placed in water

$\frac{1}{f} = [\frac{μ_{lens}}{μ_{water}} - 1] [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$
$= (\frac{1.60}{1.33} - 1) [\frac{1}{60}]$
$= \frac{28}{133 \times 60} ≃ \frac{1}{300}$
$\Rightarrow$ f = 300 cm.
Thus, the focal length of the lens is 300 cm when it is placed in water.

Page No 415:

Answer:

Given,
Refractive index of the material, (μ) = 1.50
Magnitudes of the radius of curvature:
R₁ = 20 cm and R₂ = 30 cm

From the given data we can make four possible lenses, using the lens maker formula.

$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

Four possible lenses:
(a) 1^st lens is double convex, in which R₁ = +20 cm and R₂ = −30 cm

$= 0.5 [\frac{1}{20} - \frac{1}{(- 30)}] = \frac{0.5 \times 5}{60}$
f = +24 cm

(b) 2^nd lens is double concave, in which R₁ = −20 cm and R₂ = +30 cm

$= 0.5 [\frac{- 1}{20} - \frac{1}{30}]$
f = −24 cm

(c) 3^rd lens is concave concave in which R₁ = −20 cm and R₂ = −30 cm

$= 0.5 [\frac{- 1}{20} - \frac{1}{(- 30)}]$
f = −120 cm

(d) 4^th lens is concave convex, in which R₁ = +20 cm and R₂ = +30 cm

$= 0.5 [\frac{1}{20} - \frac{1}{30}]$
f = +120 cm

Page No 415:

Answer:

Given,
A biconvex lens with two radii of curvature that have equal magnitude R.
Refractive index of the material of the lens is μ₂.
First medium of refractive index is μ₁_.
Second medium of refractive index is μ₃.
As per the question, the light beams are travelling parallel to the principal axis of the lens.
i.e., u (object distance) = ∞

(a) The light beam is incident on the lens from first medium μ₁.
Thus, refraction takes place at first surface
Using equation of refraction,

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$
Where, v is the image distance.
Applying sign convention, we get:
$\frac{μ_{2}}{v} - \frac{μ_{1}}{(- \infty)} = \frac{μ_{2} - μ_{1}}{R} \Rightarrow \frac{1}{v} = \frac{μ_{2} - μ_{1}}{μ_{2} R} \Rightarrow \Rightarrow v = \frac{μ_{2} R}{μ_{2} - μ_{1}}$

Now, refraction takes place at 2^nd surface

Thus, $\frac{μ_{3}}{v} - \frac{μ_{2}}{u} = \frac{μ_{3} - μ_{2}}{R}$
Here, the image distance of the previous case becomes object distance:

$\frac{μ_{3}}{v} = - [\frac{μ_{3} - μ_{2}}{R} - \frac{μ_{2}}{μ_{2} R} (μ_{2} - μ_{1})] = - [\frac{μ_{3} - μ_{2} - μ_{2} + μ_{1}}{R}] v = - [\frac{μ_{3} R}{μ_{3} - 2 μ_{2} + μ_{1}}]$
Therefore, the image is formed at $\frac{μ_{3} R}{2 μ_{2} - μ_{1} - μ_{3}}$

(b) The light beam is incident on the lens from second medium μ₃.
Thus, refraction takes place at second surface.
Using equation of refraction,
$\frac{μ_{2}}{v} - \frac{μ_{3}}{u} = \frac{μ_{2} - μ_{3}}{R}$
Where, v is the image distance
Applying sign convention, we get:

$\frac{μ_{2}}{v} - \frac{μ_{3}}{(- \infty)} = \frac{μ_{2} - μ_{3}}{R} \Rightarrow \frac{1}{v} = \frac{μ_{2} - μ_{3}}{μ_{2} R} \Rightarrow \Rightarrow v = \frac{μ_{2} R}{μ_{2} - μ_{3}}$
Now, refraction takes place at 2nd surface.

Thus, $\frac{μ_{1}}{v} - \frac{μ_{2}}{u} = \frac{μ_{1} - μ_{2}}{R}$

Here, the image distance of the previous case becomes object distance.
$\frac{μ_{1}}{v} = - [\frac{μ_{1} - μ_{2}}{R} - \frac{μ_{2}}{μ_{2} R} (μ_{2} - μ_{3})] = - [\frac{μ_{1} - μ_{2} - μ_{2} + μ_{3}}{R}] v = - [\frac{μ_{1} R}{μ_{3} - 2 μ_{2} + μ_{1}}]$
Therefore, the image is formed at $\frac{μ_{1} R}{2 μ_{2} - μ_{1} - μ_{3}}$

Page No 415:

Answer:

Given:
Focal length (f) of the convex lens = 10 cm

(a) As per the question, the object distance (u) is 9.8 cm.
The lens equation is given by:

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
= $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{9.8}$
= $\frac{1}{v} = \frac{9.8 - 10}{98} = \frac{- 0.2}{98}$
= v = − 98 × 5
= − 490 cm (Same side of the object)

v = 490 cm (Virtual and on on the side of object)
Magnification of the image $= \frac{v}{u}$
$= \frac{- 490}{- 9.8} = 50$
Therefore, the image is erect and virtual.

(b) Object distance, u = 10.2 cm
The lens equation is given by:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
= $\frac{1}{v} = \frac{1}{10} - \frac{1}{10.2}$
$= \frac{10.2 - 10}{102} = \frac{0.2}{102}$
= v = 102 × 5 = 510 cm (Real and on the opposite side of the object)
Magnification of the image $= \frac{v}{u}$
$= \frac{510}{- 9.8} = - 52.04$
Therefore, the image is real and inverted.

Page No 415:

Answer:

Given,
We are projecting a slide of 35 mm $\times$ 23 mm on a 2 m $\times$ 2 m screen using projector.
Therefore, the magnification required by the projector is:
$m = \frac{v}{u}$
Here,
v = Image distance
u = Object distance
We will take 35 mm as the object size
∵ 35 mm > 23 mm
$∴ m = \frac{v}{u} = \frac{h_{i}}{h_{o}} \frac{v}{u} = \frac{2 \times 10^{3}}{35} \Rightarrow u = \frac{35}{2 \times 10^{3}} \times v \Rightarrow u = \frac{35}{2 \times 10^{3}} \times 10 \Rightarrow u = 0.175 mm$

The lens formula is given by

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\Rightarrow \frac{1}{10} + \frac{1}{0.175} = \frac{1}{f}$

$\Rightarrow \frac{10175}{1750} = \frac{1}{f}$

$\Rightarrow f = \frac{1750}{10175} f = 0.172 mm$

Hence, the required focal length is 0.172 mm.

Page No 415:

Answer:

When the particle is at point B,

$\frac{1}{v_{B}} = \frac{1}{f} + \frac{1}{u}$
$\frac{1}{v_{B}} = \frac{1}{12} - \frac{1}{19}$
$\Rightarrow v_{B} = \frac{12 \times 19}{7} \Rightarrow v_{B} = 32.57 cm$

When particle is at point A,

$\frac{1}{v_{A}} = \frac{1}{f} + \frac{1}{u}$
$\frac{1}{v_{A}} = \frac{1}{12} - \frac{1}{21}$
$\Rightarrow v_{A} = \frac{12 \times 21}{9} \Rightarrow v_{A} = 28 cm$

$Amplitude of image = \frac{v_{A} - v_{B}}{2}$
$= \frac{4.5}{2} = 2.2 cm$

Page No 415:

Answer:

Given,
Object distance, (u) = 5.0 cm
Focal length (f) of convex lens = 8.0 cm

(a)

(b)
Using lens formula:

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Where v is the image distance,
on putting the given values we get:

$\frac{1}{v} - \frac{1}{(- 5)} = \frac{1}{8} \Rightarrow \frac{1}{v} = \frac{1}{8} - \frac{1}{5} \Rightarrow \frac{1}{v} = \frac{- 3}{40} ∴ v = - 13.3 cm$

Hence, the position of the image from the lens is 13.3 cm.

Page No 415:

Answer:

Given,
Height of the object, (h₀) = 2 cm,
Height of the image, (h₁) = 1 cm
distance between image and object, ( $-$ u + v) = 40 cm

We know that,
Magnification is given by:
$m = \frac{h_{i}}{h_{0}} = \frac{v}{u}, \frac{1}{2} = \frac{v}{- u}$
u = −2v

Using lens formula,

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} - \frac{1}{- 2 v} = \frac{1}{f} \Rightarrow \frac{1}{v} + \frac{1}{2 v} = \frac{1}{f} \Rightarrow f = \frac{2 v}{3}$
As per the question,
= −u + v = 40 cm,
= −(−2v) + v = 40 cm,
= 3v = 40 cm

$v = 13.3 = \frac{40}{3} cm, u = \frac{80}{3}$ , u = 26.66 cm.
∴ f = 8.9 cm

Hence, the required focal length is 8.9 cm and object distance is 26.66 cm.

Page No 416:

Answer:

Given,
Object distance, u = $-$ 18 cm
Size of the image is two times the size of the object
i.e., h₁ = 2h₀,
We know that,
magnification is given by, m = $\frac{v}{u}$
$\Rightarrow \frac{h_{1}}{h_{0}} = \frac{v}{u} = 2$
∴ v = 2u = 36 cm.

Using lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{36} + \frac{1}{18} = \frac{1}{f}$

$\Rightarrow \frac{1}{f} = \frac{3}{36}$

⇒ f = 12 cm.

For, h₁ = 3h₀

$\frac{h_{i}}{h_{0}} = \frac{v}{u}$

$\Rightarrow \frac{v}{u} = 3 \Rightarrow v = 3 u$

$- \frac{1}{3 u} - \frac{1}{u} = \frac{1}{12}$

u = $-$ 16 cm.

Therefore, the object should be placed at a distance of 16 cm in front of the lens.

Page No 416:

Answer:

Given,
Length of the pin = 2.0 cm
Focal length (f) of the lens = 6 cm
As per the question, the centre of the pin is 11 cm away from the lens.
i.e., the object distance (u) = 10 cm

Since, we have to calculate the image of A and B, Let the image be A' and B'
So, the length of the A'B' = size of the image.

Using lens formula:
$\frac{1}{v_{A}} - \frac{1}{u_{A}} = \frac{1}{f}$
Where v_A and u_A are the image and object distances from point A.
$\Rightarrow \frac{1}{v_{A}} - \frac{1}{- 10} = \frac{1}{6}$
$\frac{1}{v_{A}} = \frac{1}{6} - \frac{1}{10} = \frac{1}{15}$
$v_{A} = 15 cm$

Similarly for point B,
Lens formula:

$\frac{1}{v_{B}} - \frac{1}{u_{B}} = \frac{1}{f}$
Where v_A and u_A are the image and object distances from point B.
$\frac{1}{v_{B}} - \frac{1}{- 12} = \frac{1}{6} \Rightarrow v_{B} = 12 cm$
Length of image = v_A − v_B = 15 − 12 = 3 cm.

Page No 416:

Answer:

Given,
Diameter of the sun = 1.4 × 10⁹ m
Distance between sun and earth is taken as object distance (u) = − 150 × 10¹¹ cm,
Focal length (f) of the lens = 20 cm
Using lens formula,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
The object distance is very large as compared to focal length of the lens.
Hence, the image is formed at the focus.

$\Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{150 \times 10^{11}}$

$= \frac{750 \times 10^{9} - 1}{150 \times 10^{11}}$

$≃ \frac{750 \times 10^{9}}{150 \times 10^{11}}$

$\Rightarrow v = \frac{150 \times 10^{11}}{750 \times 10^{9}}$

We know, Magnification (m) is given by:

$(m) = \frac{v}{u} = \frac{h_{2}}{h_{1}}$

$\Rightarrow h_{2} = \frac{v}{u} \times h_{1}$

$= \frac{150 \times 10^{11}}{750 \times 10^{9} \times 150 \times 10^{11}} \times 0.7 \times 10^{12} mm = 0.93 mm$

Hence, the required radius of the image of the sun is 0.93 mm.

Page No 416:

Answer:

Given,
Power of the lens (P) = $\frac{1}{Focal length}$ = 5.0 D
The height of the image is four times the height of the object.
i.e.
$h_{i} = 4 h_{0}, \frac{h_{i}}{h_{0}} = 4$
We know magnification (m) is also given by
$m = \frac{h_{i}}{h_{0}} = \frac{v}{u} \Rightarrow \frac{v}{u} = 4, v = 4 u$
The lens maker formula is given by

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Here, v is the image distance and u is the object distance.
Now,
$\frac{1}{f} = P = 5 and f = \frac{1}{5} m = 20 cm \frac{1}{4 u} - \frac{1}{u} = \frac{1}{20} - \frac{3}{4 u} = \frac{1}{20} \Rightarrow u = - \frac{60}{4} = - 15 cm$

Hence, the required object distance is 15 cm.

Page No 416:

Answer:

Let the object be placed at a distance x cm from the lens (away from the mirror).
For the concave lens (I^st refraction) u = − x, f = − 20 cm
From lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{(- 20)} + \frac{1}{(- x)} \Rightarrow v = - (\frac{20 x}{x + 20})$
Thus, the virtual image due to the first refraction lies on the same side as that of object (A'B').
This image becomes the object for the concave mirror,
For the mirror,
$u = - (5 + \frac{20 x}{x + 20}) = - (\frac{25 x + 100}{x + 20}) f = - 10 cm$

From mirror equation,
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{- 10} + \frac{x + 20}{25 x + 100} \Rightarrow \frac{1}{v} = \frac{10 x + 200 - 25 x - 100}{250 (x + 4)}$
$\Rightarrow v = \frac{250 (x + 4)}{100 - 15 x} \Rightarrow v = \frac{250 (x + 4)}{15 x - 100} \Rightarrow v = \frac{50 (x + 4)}{(3 x - 20)}$
Thus, this image is formed towards left of the mirror.

Again for second refraction in concave lens,
$u = - [\frac{5 - 50 (x + 4)}{3 x - 20}]$
(assuming that image of mirror is formed between the lens and mirror 3x − 20),
v = + x (since the final image is produced on the object A"B")
using lens formula,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{x} + \frac{1}{\frac{[5 - 50 (x \times 4)]}{3 x - 20}} = \frac{1}{- 20}$
⇒ 25x² − 1400x − 6000 = 0
⇒          x² − 56x − 240 = 0
⇒          (x − 60) (x + 4) = 0
So,                           x = 60 m
The object should be placed at a distance 60 cm from the lens farther away from the mirror, so that the final image is formed on itself.

Page No 416:

Answer:

Let the object be placed at a distance x cm from the lens (away from the mirror).
For the convex lens (1^st refraction) u = − x, f = − 12 cm

From the lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{(- 12)} + \frac{1}{(- x)} \Rightarrow v = - (\frac{12 x}{x + 12})$
Thus, the virtual image due to the first refraction lies on the same side as that of object A'B'.
This image becomes the object for the convex mirror,
For the mirror,
$u = - (5 + \frac{12 x}{x + 12}) = - (\frac{17 x + 60}{x + 12}) f = - 7.5 cm$

From mirror equation,
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{- 7.5} + \frac{x + 12}{17 x + 60} \Rightarrow \frac{1}{v} = \frac{17 x + 60 - 7.5}{7.5 (17 x + 60)}$
$\Rightarrow v = \frac{7.5 (17 x + 60)}{52.5 - 127.5 x} \Rightarrow v = \frac{250 (x + 4)}{15 x - 100} \Rightarrow v = \frac{50 (x + 4)}{(3 x - 20)}$

Thus, this image is formed towards the left of the mirror.
Again for second refraction in concave lens,
$u = - [\frac{5 - 50 (x + 4)}{3 x - 20}]$
(assuming that the image of mirror formed between the lens and mirror is 3x − 20),
v = + x (since, the final image is produced on the object A"B")
Using lens formula:

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{x} + \frac{1}{\frac{[5 - 50 (x \times 4)]}{3 x - 20}} = \frac{1}{- 20}$
⇒ 25x² − 1400x − 6000 = 0
⇒          x² − 56x − 240 = 0
⇒          (x − 60) (x + 4) = 0
Thus,                           x = 60 m
The object should be placed at a distance 60 cm from the lens farther away from the mirror, so that the final image is formed on itself.

Page No 416:

Answer:

Given,
Distance between convex lens and convex mirror is 15 cm.
Focal length (f_l) of the lens is 25 cm.
Focal length (f₂) of the mirror is 40 cm.
Let x cm be the object distance from the mirror.
Therefore,
u = − x cm
v = 25 − 15 = + 10 cm (∵ focal length of lens = 25 cm)
∴ f_l= + 40 cm

Using lens formula:

$\Rightarrow \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{x} = \frac{1}{10} - \frac{1}{40} \Rightarrow x = \frac{40}{3}$
Thus, the object distance is $(15 - \frac{40}{3}) = \frac{5}{3}$
= 1.67 cm from the lens

Page No 416:

Answer:

Given:
Convex lens of focal length (f_l) = 15 cm
Concave mirror of focal length (f₂) = 10 cm
Distance between the lens and the mirror = 50 cm
Point source is placed at a distance of 40 cm from the lens.
It means the point source is at the focus of the mirror.
Thus, two images will be formed:
(a) One due to direct transmission of light through the lens.
(b) One due to reflection and then transmission of the rays through the lens.

Case 1:
(S') For the image by direct transmission, we have:
Object distance (u) = − 40 cm
f_l = 15 cm
Using the lens formula, we get:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{15} + \frac{1}{(- 40)} = \frac{40 - 15}{40 \times 15} \Rightarrow v = \frac{40 \times 15}{40 - 15}$

Therefore, v is 24 cm to the left from the lens.

Case II:
(S') Since the object is placed at the focus of the mirror, the rays become parallel to the lens after reflection.
∴ Object distance (u) = ∞ ⇒ f_l = 15 cm
$\Rightarrow \frac{1}{v} - \frac{1}{u} = \frac{1}{15}$
Thus, v is 15 cm to the left of the lens.

Page No 416:

Answer:

Given,
Convex lens of focal length (f_l) = 15 cm
Concave mirror of focal length (f_m) = 10 cm
Distance between lens and mirror = 50 cm
Thus, two images will be formed,
(a) One due to direct transmission of light through lens.
(b) One due to reflection and then transmission of the rays through lens.

Let the point source be placed at a distance of 'x' from the lens as shown in the Figure, so that images formed by lens and mirror coincide.
For lens,
We use lens formula:

$\frac{1}{v} - \frac{1}{u} = \frac{1}{15} \Rightarrow v = \frac{15 x}{x - 15} . . . (i)$
For mirror,
Object distance will be (50 − x)
We use the formula:

$\frac{1}{v} + \frac{1}{u} = \frac{1}{15}$
u = − (50 − x)
f_m = − 10 cm
$So, \frac{1}{v_{m}} + \frac{1}{- (50 - x)} = - \frac{1}{10} \Rightarrow \frac{1}{v_{m}} = \frac{1}{50 - x} - \frac{1}{10} = \frac{10 - 50 + x}{10 (50 - x)} v_{m} = \frac{x - 40}{10 (50 - x)} . . . (ii)$

Since the distance between lens and mirror is 50 cm,
v − v_m = 50
from equation (i) and (ii):

$\Rightarrow \frac{15 x}{x - 15} - \frac{10 (50 - x)}{(x - 40)} = 50$
⇒(3x² − 120x) − (100x − 2x² − 1500 + 30x)
                                 = 10 (x² − 55x + 600)
⇒ 5x²− 250x − 1500 = 10x² − 550x + 6000
⇒ 5x² − 300x + 4500 = 0
⇒       x² − 60x + 900 = 0
⇒                  (x − 30)² = 0
x = 30 cm.
∴ Point source should be placed at a distance of 30 cm from the lens on the principal axis, so that the two images form at the same place.

Page No 416:

Answer:

Given,
Convex lens of focal length f_l = 15 cm
Concave mirror of focal length f_m = 10 cm
Distance between mirror and lens = 50 cm
Length of the pin (object length) h₀ = 2.0 cm

As per the question
The pin (object) is placed at a distance of 30 cm from the lens on the principle axis.
Using lens formula,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f_{l}} \Rightarrow v = \frac{u f_{l}}{u + f_{l}}$

Since, u = −30 cm and f_l = 15 cm
$So, v = \frac{(- 30) \times 15}{(- 30 + 15)} = \frac{- 450}{- 15} = 30 cm$
From the figure it can be seen that image of the object (AB) is real and inverted (A'B') and it is of the same size as the object. This image (A'B') is at a distance of 20 cm from the concave mirror, which is formed at the centre of curvature of the mirror. Thus, mirror will form the image (A'B') at the same place as (A''B'') and will be of the same size. Now, due to the refraction from the lens, the final image (A''B'') will be formed at AB and will be of the same size as the object (AB).

Page No 416:

Answer:

Given,
Convex lens of focal length (f) = 15 cm
Object distance, (u) = −30 cm
Using lens formula,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \Rightarrow \frac{1}{v} = \frac{1}{15} - \frac{1}{30} = \frac{1}{30} \Rightarrow v = 30$

Thus ,v = 30 cm
Therefore, the image of the object will be formed at a distance of 30 cm to the right side of the lens.
[since, μ_g = 1.5 and thickness (t) = 1 cm]

$Δ t = (1 - \frac{1}{μ_{g}}) t$
$= 1 - \frac{2}{3} = \frac{1}{3} = 0.33 cm$
Hence, the image of the object will be formed at 30 + 0.33 = 30.33 cm from the lens on the right side, due to the glass having thickness 1 cm.

Page No 416:

Answer:

Given,
Focal length of the convex lens, f_d = 20 cm
Focal length of the concave lens, f_c = 10 cm
Beam diameter of the incident light, d = 5.0 mm
Distance between both the lenses is 10 cm.

As per the question, the incident beam of light is parallel to the principal axis.
Let it be incident on the convex lens.
Now, let B be the focus of the convex lens where the image by the convex lens should be formed.
For the concave lens,
Object distance (u) = + 10 cm (Virtual object is on the right of concave lens.)
Focal length, f_c = − 10 cm
Using the lens formula,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{- 10} + \frac{1}{+ 10} = \infty \Rightarrow v = \infty$
Thus, after refraction in the concave lens, the emergent beam becomes parallel.
As shown, in triangles XYB and PQB,

$\frac{PQ}{XY} = \frac{RB}{ZB} = \frac{10}{20} = \frac{1}{2} PQ = \frac{1}{2} \times 5 = 2.5 mm$

Thus, the beam diameter of the emergent light is 2.5 mm.
Similarly, we can prove that if the beam of light is incident on the side of the concave lens, the beam diameter (d) of the emergent light will be 1 cm.

Page No 416:

Answer:

Given,
Focal length of convex lens, f_c = 30 cm
Focal length of concave lens, f_d = 15 cm
Distance between both the lenses, d = 15 cm
Let (f) be the equivalent focal length of both the lenses.
$\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{d}{f_{1} f_{2}} = \frac{1}{30} + (\frac{1}{20}) - (\frac{5}{30 (- 20)}) = \frac{1}{120} \Rightarrow f = 120 cm$
As focal length is positive, so it will be a converging lens.
Let 'd₁' be the distance from diverging lens, so that the emergent beam is parallel to the principal axis and the image will be formed at infinity.
$d_{1} = \frac{d f}{f_{c}} = \frac{15 \times 120}{30} = 60 cm$
It should be placed 60 cm left to the diverging lens. The object should be placed
(120 − 60) = 60 cm from the diverging lens
Let d₂ be the distance from the converging lens. Then,
$d_{2} = \frac{d f}{f_{d}} = \frac{15 \times 120}{20}$
d₂ = 90 cm
Thus, it should be placed (120 + 90) cm = 210 cm right to converging lens.

Page No 416:

Answer:

Given:
Length of the high pin = 5.00 mm
Focal length of the first convex lens, f = 10 cm
Distance between the first lens and the pin = 15 cm
Focal length of the second convex lens, f₁ = 5 cm
Distance between the first lens and the second lens = 40 cm
Distance between the second lens and the pin = 55 cm

(a) Image formed by the first lens:
Here,
Object distance, u = − 15 cm
Focal length, f = 10 cm
The lens formula is given by

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \frac{1}{v} = \frac{1}{10} - \frac{1}{15} \Rightarrow v = 30 cm$

Now,
This will be object for the second lens.
∴ Object distance for the second lens, $u_{1}$ = − (40 − 30)
$\Rightarrow u_{1}$ = − 10 cm
Focal length of the second lens, f₁= 5 cm
The lens formula is given by

$\frac{1}{v_{1}} = \frac{1}{f_{1}} + \frac{1}{u_{1}} \Rightarrow \frac{1}{v_{1}} = \frac{1}{5} - \frac{1}{10} \Rightarrow v_{1} = 10 cm$
Therefore, the final position of the image is 10 cm right from the second lens.

(b) Magnification $(m)$ by the first lens is given by
$m = \frac{h_{i}}{h_{0}} = \frac{v}{u} \Rightarrow h_{i} = - \frac{5 \times 30}{15} \Rightarrow h_{i} = - 10 mm$
Magnification by the second lens:

$\frac{h_{final}}{h_{i}} = \frac{v}{u} \Rightarrow \frac{10}{- 10} = \frac{h_{final}}{- 10} \Rightarrow h_{final} = 10 mm$
Thus, the image will be erect and real.
(c) Size of the final image is 10 mm.

Page No 416:

Answer:

Given,
Distance between point object and convex lens, u = 15 cm
Distance between the image of the point object and convex lens, v = 30 cm
Let f_c be the focal length of the convex lens.

Then, using lens formula, we have:

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f_{c}} \Rightarrow \frac{1}{f_{c}} = \frac{1}{30} - \frac{1}{(- 15)} \Rightarrow \frac{1}{f_{c}} = \frac{1}{30} + \frac{1}{15} = \frac{3}{30} \Rightarrow f_{c} = 10 cm$
Now, as per the question, the concave lens is placed in contact with the convex lens. So the image is shifted by a distance of 30 cm.
Again, let v_f be the final image distance from concave lens, then:
$v_{f}$ = + (30 + 30) = + 60 cm
Object distance from the concave lens, v = 30 cm
If f_d is the focal length of concave lens then
Using lens formula, we have:

$\frac{1}{v_{f}} - \frac{1}{v} = \frac{1}{f_{d}} \Rightarrow \frac{1}{f_{d}} = \frac{1}{60} - \frac{1}{30} \Rightarrow \frac{1}{f_{d}} = \frac{30 - 60}{60 \times 30} = \frac{- 30}{60 \times 30} \Rightarrow f_{d} = - 60 cm$
Hence, the focal length (f_c) of convex lens is 10 cm and that of the concave lens (f_d ) is 60 cm.

Page No 416:

Answer:

Given,
Focal length of each convex lens, f = 10 cm
Distance between both the lens, d = 15 cm

(a) As shown in the figure, the light rays are falling parallel to the principal axis of the first lens, therefore the rays will converge within the focal length of the second lens. Hence, all the rays emerging from the lens system are diverging.
(b) Using lens formula for first convex lens,
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} \Rightarrow \frac{1}{v} = \frac{1}{10} + \frac{1}{(- \infty)} \Rightarrow \frac{1}{v} = \frac{1}{10} \Rightarrow v = 10 cm$

For the second convex lens,
Object distance, u = −(15 − 10) = −5 cm
If v₁ is the image distance for the second convex lens, then,
Applying lens formula we have:

$\frac{1}{v_{1}} = \frac{1}{f} + \frac{1}{u}$
$\Rightarrow \frac{1}{v_{1}} = \frac{1}{10} + \frac{1}{- 5} \Rightarrow \frac{1}{v_{1}} = \frac{- 1}{10} \Rightarrow v_{1} = - 10 cm$
Thus, the virtual image of the object will be at 5 cm from the first convex lens.

(c) The equivalent focal length of the lens system ^$(F)$ is given by,

$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{d}{f_{1} f_{2}} Here, f = f_{1} = f_{2} = 10 cm ∴ \frac{1}{F} = \frac{1}{10} + \frac{1}{10} - \frac{15}{100} \Rightarrow \frac{1}{F} = \frac{5}{100} \Rightarrow F = 20 cm$

Therefore the equivalent focal length (F) is 20 cm

Page No 416:

Answer:

Given,
A transparent sphere with refractive index μ and a of radius R.
A ball is kept at height h above the sphere.
As per the question, at t = 0 the ball is dropped normally on the sphere.

Let take the time taken to travel from point A to B as t .
Thus, the distance travelled by the ball is given by:
= $\frac{1}{2} g t^{2}$
Where g is acceleration due to gravity.
Therefore, the distance BC is given by:
$= h - \frac{1}{2} g t^{2}$
We are assuming this distance of the object from lens at any time t.
So here,
u = $- (h - \frac{1}{2} g t^{2})$
Taking:
Refractive index of air, μ₁ = 1
Refractive index of sphere, μ₂ = μ (given)
Thus,
$\frac{μ}{v} - \frac{1}{- (h - \frac{1}{2} g t^{2})} = \frac{μ - 1}{R} \Rightarrow \frac{μ}{v} = \frac{μ - 1}{R} - \frac{1}{(h - \frac{1}{2} g t^{2})} = \frac{(μ - 1) (h - \frac{1}{2} g t^{2}) - R}{R (h - \frac{1}{2} g t^{2})}$
Let v be the image distance at any time t. Then,
$v = \frac{μ R (h - \frac{1}{2} g t^{2})}{(μ - 1) (h - \frac{1}{2} g t^{2}) - R}$

Therefore, velocity of the image $(V)$ is given by,
$V = \frac{d v}{d t} = \frac{d}{d t} [\frac{μ R (h - \frac{1}{2} g t^{2})}{(μ - 1) (h - \frac{1}{2} g t^{2}) - R}] = \frac{μ R^{2} g t}{{[(μ - 1) (h - \frac{1}{2} g t^{2}) - R]}^{2}}$

Page No 416:

Answer:

Given,
Radius of the concave mirror is R
Therefore focal length of the mirror, $f = \frac{R}{2}$
Velocity of the particle, $V = \frac{d x}{d t}$
Object distance, u = −x

Using mirror equation,
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
On putting the respective values we get,
$\frac{1}{v} + \frac{1}{- x} = - \frac{2}{R} \Rightarrow \frac{1}{v} = - \frac{2}{R} + \frac{1}{x} = \frac{R - 2 x}{R x} ∴ v = \frac{R x}{R - 2 x}$

Velocity of the image is given by V₁
$V_{1} = \frac{dv}{d t} = \frac{d}{d t} [\frac{R x}{R - 2 x}] = \frac{[\frac{d}{d t} (R x) (R - 2 x)] - [\frac{d}{d t} (R - 2 x) (R x)]}{{(R - 2 x)}^{2}} = \frac{R [(\frac{dx}{d t}) (R - 2 x)] - [2 \frac{dx}{d t} x]}{{(R - 2 x)}^{2}} = \frac{R [(V) (R - 2 x)] - [2 V \times 0]}{{(R - 2 x)}^{2}} = \frac{V R^{2}}{(2 x - R)^{2}}$

Page No 416:

Answer:

Note :
(a) At time t = t,
   Object distance, u = −(d − Vt)
   Here d > Vt, $f = - \frac{R}{2}$

Mirror formula is given by:

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$\Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = - \frac{2}{R} + \frac{1}{d - V t}$
$= \frac{- 2 (d - V t) + R}{R (d - V t)}$
$v = \frac{- R (d - V t)}{R - 2 (d - V t)}$

Differentiating w.r.t. 't':

$\frac{d v}{d t} = \frac{- RV [R - 2 (d - V t)] - 2 V [R (d - V t)]}{{[R - 2 (d - V t)]}^{2}} = - \frac{R^{2} V}{{[2 (d - V t) - R]}^{2}}$
This is the required speed of mirror.
(b) When $t > \frac{d}{2}$ the collision between the mirror and mass will take place. As the collision is elastic, the object will come to rest and the mirror will start to move with the velocity V.

u = (d − Vt).
At any time, $t > \frac{d}{2}$
The distance of mirror from the mass will be:
$x = V (t - \frac{d}{V}) = V t - d$
Here,
Object distance, u = $- (V t - d)$ = $(d - V t)$
Focal length, f = $- \frac{R}{2}$
Now, mirror formula is given by:
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$\Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ = $\frac{1}{- \frac{R}{2}} - \frac{1}{d - V t}$
$v \Rightarrow - [\frac{R (d - V t)}{R - 2 (d - V t)}]$

Velocity of image $(V_{i m a g e})$ is given by,
$V_{i m a g e} = \frac{d v}{dt} = \frac{d}{dt} [\frac{R (d - V t)}{R + 2 (d - V t)}]$
If y = d $-$ Vt
$\frac{d y}{dt}$ = $- V$

Velocity of image:

$V_{i m a g e} = \frac{d}{d t} [\frac{R y}{R + 2 y}] = - V_{r} [\frac{R + 2 y - 2 y}{{(R + 2 y)}^{2}}] = \frac{- V R^{2}}{{(R + 2 y)}^{2}}$
As the mirror is moving with velocity V,
Therefore,
V=V_image + V_mirror
Absolute velocity of image = $V [1 - \frac{R^{2}}{{(R + 2 y)}^{2}}]$
                                        = $V [1 - \frac{R^{2}}{2 (V t - d) - R^{2}}]$

Page No 416:

Answer:

Given,
The focal length of the concave mirror is f and M is the mass of the gun. Horizontal speed of the bullet is V.

Let the recoil speed of the gun be V_g
Using the conservation of linear momentum we can write,
MV_g = mV
$\Rightarrow V_{g} = \frac{m}{M} V$
Considering the position of gun and bullet at time t = t,
For the mirror, object distance, u = − (Vt + V_gt)
Focal length, f = − f
Image distance, v = ?
Using Mirror formula, we have:

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{- f} - \frac{1}{u} \Rightarrow \frac{1}{v} = - \frac{1}{f} + \frac{1}{(V t + V_{g} t)} \Rightarrow \frac{1}{v} = \frac{- (V t + V_{g} t) + f}{(V t + V_{g} t) f} \Rightarrow v = \frac{- (V + V_{g}) t f}{f - (V + V_{g}) t}$

The separation between image of the bullet and bullet at time t is given by:

$v - u = \frac{- (V + V_{g}) t f}{f - (V + V g) t} + (V + V_{g}) t = (V + V_{g}) t [\frac{f}{f - (V + V_{g}) t} + 1] = 2 (1 + \frac{m}{M}) V t$

Differentiating the above equation with respect to 't' we get,
$\frac{d (v - u)}{d t} = 2 (1 + \frac{m}{M}) V$
Therefore, the speed of separation of the bullet and image just after the gun was fired is $2 (1 + \frac{m}{M}) V$ .

Page No 416:

Answer:

Given,
Mass = 50 g
Spring constant, k = 500 Nm⁻¹
Height from where the mass is dropped on the spring, h = 10 cm
Focal length of concave mirror, f = 12 cm
Distance between the pole and the free end of the spring is 30 cm.
As per the question, when the mass is released it will stick to the spring and execute SHM.
At equilibrium position, weight of the mass is equal to force applied by the spring.
∴ mg = kx
where g is acceleration due to gravity
⇒ x = mg/k
$= \frac{50 \times 10^{- 3} \times 10}{500} = 10^{- 3} m = 0.1 cm$

Therefore, the mean position of the SHM is (30 + 0.1 = 30.1) cm away from the pole of the mirror.
From the work energy principle,
final kinetic energy − initial kinetic energy = work done
⇒ 0 − 0 = mg(h + δ) − $\frac{1}{2} k δ^{2}$
Where δ is the maximum compression of the spring.
mg(h + δ) = $\frac{1}{2} k δ^{2}$
$\Rightarrow 50 \times 10^{- 3} \times 10 (0.1 + δ) = \frac{1}{2} 500 δ^{2} δ = \frac{0.5 \pm \sqrt{0.25 + 50}}{2 \times 250} = 0.015 m = 1.5 cm$

From the above figure,
Position of point B, (30 + 1.5) = 31.5 cm from the pole of the mirror
Therefore, amplitude of vibration of SHM, (31.5 − 30.1) = 1.4 cm
Position of point A from the pole of the mirror, (30.1 − 30.1) = 28.7 cm
For point A,
Object distance (u_a) = − 31.5
f = − 12 cm
By using the lens formula:

$\frac{1}{v_{a}} - \frac{1}{u_{a}} = \frac{1}{f} \Rightarrow \frac{1}{v_{a}} = \frac{1}{f} + \frac{1}{u_{a}} = \frac{1}{- 12} + \frac{1}{- 31.5} = v_{a} = - 19.38 cm$

For point B,
Object distance, (u_b) = − 28.7
f = − 12 cm
By using the lens formula:

$\frac{1}{v_{b}} - \frac{1}{u_{b}} = \frac{1}{f} \Rightarrow \frac{1}{v_{b}} = \frac{1}{f} + \frac{1}{u_{b}} = \frac{1}{- 12} + \frac{1}{- 28.7} = v_{b} = - 20.62 cm$
Hence, the image of the mass oscillates in length (20.62 − 19.38) = 1.24 cm

Page No 417:

Answer:

Given,
R is the radii of curvature of two concave mirrors and M is the mass of the whole system.
Mass of the two blocks A and B is m.
As per the question,
At t = 0,
distance between block A and B is 2R
Block B is moving at a speed v towards the mirror.
Original position of the whole system at x = 0
(a)

At time t = $\frac{R}{v}$
The block B moved $(v \times \frac{R}{v} = R)$ R distance towards the mirror.
For block A,
object distance, u = − 2R
focal length of the mirror, f = − $\frac{R}{2}$
Using the mirror formula:
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = - \frac{2}{R} + \frac{1}{2 R} = - \frac{3}{2 R}$
Therefore, v = − $\frac{2 R}{3}$
Position of the image of block A is at $\frac{2 R}{3}$ with respect to the given coordinate system.
For block B,
Object distance, u = − R
Focal length of the mirror, f = − $\frac{R}{2}$
Using the mirror formula:
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \Rightarrow \frac{1}{v} = - \frac{2}{R} + \frac{1}{R} \Rightarrow \frac{1}{v} = - \frac{1}{R}$
Therefore, v = − R
Position of the image of block B is at the same place.
Similarly,
(b)

At time $\frac{3 R}{v}$
Block B, after colliding with the mirror must have come to rest because the collision is elastic. Due to this, the mirror has travelled a distance R towards the block A, i.e., towards left from its initial position.
So, at this time
For block A
Object distance, u = − R
Focal length of the mirror, f = − $\frac{R}{2}$
Using the mirror formula:
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = - \frac{2}{R} + \frac{1}{R} = - \frac{1}{R}$
Therefore, v = − R
Position of the image of block A is at − 2R with respect to the given coordinate system.

For Block B,
Image of the block B is at the same place as it is at a distance of R from the mirror.
Therefore, the image of the block B is zero with respect to the given coordinate system.
(c)

At time $\frac{5 R}{v}$
In a similar manner, we can prove that the position of the image of block A and B will be at − 3R and $\frac{- 4 R}{3}$ respectively.

Page No 417:

Answer:

Given,
Acceleration of the elevator, a = 2.00 m/s²
Focal length of the mirror M, f = 12.00 cm
Acceleration due to gravity, g = 10 m/s²
Mass of blocks A and B = m
As per the question, the mass–pulley system is released at time t = 0.
Let the acceleration of the masses A and B with respect to the elevator be a.

Using the free body diagram,
T − mg + ma − 2m = 0 ...(i)
Also,
T − ma = 0 ...(ii)
From (i) and (ii), we get:
2ma = m(g + 2)
$\Rightarrow a = \frac{10 + 2}{2} = \frac{12}{2} = 6 {ms}^{- 2}$
Now, the distance travelled by block B of mass m in time t = 0.2 s is given by
$s = u t + \frac{1}{2} a t^{2} As (u = 0) s = \frac{1}{2} a t^{2}$
On putting the respective values, we get:
$s = \frac{1}{2} \times 6 \times {(0.2)}^{2} = 0.12 m = 12 cm$

As given in the question, the distance of block B from the mirror is 42 cm.
Object distance u from the mirror = − (42 − 12) = − 30 cm
Using the mirror equation,
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
On putting the respective values, we get:
$\frac{1}{v} + \frac{1}{- 30} = \frac{1}{12} \Rightarrow \frac{1}{v} = \frac{1}{- 30} = \frac{1}{12} + \frac{1}{30} \Rightarrow v = 8.57 cm$
Hence, the distance between block B and mirror M is 8.57 cm.

View NCERT Solutions for all chapters of Class 11

Login or Create a free account