Hc Verma I for Class 11 Science Physics Chapter 19 - Optical Instruments

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The retina acts as a screen; only real images can be obtained on the screen. In case of people having eye defects, the spectacles form the virtual image of the object and the eye lens form the real and inverted image on the retina.

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The image formed by a simple microscope is virtual and erect. So, it cannot be projected on a screen without using any additional lens or mirror.

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No, angular magnification is the ratio of the angle subtended by the final image on the eye to the angle subtended by the object on the unaided eye. Its value less than one signifies reduction in the size of the image. It does not mean that the image is inverted.

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A simple microscope consists of a single convex lens. Sometimes due to chromatic and spherical aberrations, the image of a white source seems coloured at the corners of the lens and somewhere in between.

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A converging lens in a magnifying glass is of small focal length which is used to magnify an object which is placed close to the lens. On the other hand, converging lens used as spectacles is of varying focal length which depends upon the actual near point of the long-sighted person. It forms image at the near point of the defected eye which is further focussed by the eye lens.

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If a person views an extended object through a converging lens, then it will appear larger and wider to him.

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In a simple microscope, the converging lens is used to magnify the object. It is done by the eyepiece in a compound microscope. But the purpose of the objective lens is the same, i.e., to form an enlarged, real and inverted image of the object at a distance less than the focal length of the eyepiece. So, its magnification power cannot be expressed in a way it is expressed for a simple microscope.

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The image formed by a concave lens is virtual and upright. It is smaller than the object and is formed between the object and the lens, irrespective of the position of the object. If, by mistake, an eye surgeon puts a concave lens in place of eye lens in a patient's eye, then the image will not be focused on the retina; this will lead to unclear vision.

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The magnifying power of a simple microscope depends on the ratio $\frac{D}{f}$ for a farsighted person. Here, D for a farsighted person is greater than that for a normal person, but the value of f remains the same. Therefore, the magnifying power of a simple microscope is greater for a farsighted person compared to that for a person with normal vision. Also, a farsighted person can see the insect more clearly under the microscope than a person with normal vision.

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Instruments like telescopes and microscopes deal with objects placed at different distances. Due to some physical factors, there is a relative change in heights not in the angle which the light emerging from them subtends on the lens. So, the magnification properties of instruments are defined in terms of the ratio of angles.

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Object distance, u = $-$ 30 cm
Focal length, f = 15 cm
Image distance, v = ?
The lens formula is given by
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} - \frac{1}{- 30} = \frac{1}{15} \Rightarrow \frac{1}{v} = \frac{1}{30} \Rightarrow v = + 30 cm$
(on the opposite side of the object)

(a) No, the eye placed close to the lens cannot see the object clearly.

(b) The eye should be 30 cm away from the lens to see the object clearly.

(c) The diverging lens will always form an image at a large distance from the eye; this image cannot be seen through the human eye.

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(d) size of the image formed on the retina

An eye consists of a lens that works on the principle on which a glass lens works. It forms the image on the screen called retina. The magnification, in this case, depends on the ratio of the image to the object height.

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(a) far away from the eye

A normal eye can see from 25 cm to infinity; it faces least difficulty and strain focussing on the object as far as it could be.

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(d) the eye is not able to decrease the focal length beyond a limit

The ciliary muscles adjust the focal length to form an image on the retina, but the muscles cannot be strained beyond a limit. Hence, if the object is brought too close to the eye, the focal length cannot be adjusted to form the image on the retina.

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(d) The image distance from the eye lens

In the human eye, the image is formed on the retina, which is at a fixed distance from the eye lens.

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(b) 18 cm to 200 cm

Person B has stronger ciliary muscles than person A. So, the muscles in his case can be strained more and the focal length of his eye can be reduced more compared to those of person A. While seeing far objects, the muscles are relaxed, so their strength will not affect the far point of the eye.

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(b) 2 cm

Given:
Near point of the human eye, u = $-$ 25 cm
Distance between the retina and the eye lens, v = 2 cm (approximately)
Thus, we have the focal length, f.
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{f} ≅ \frac{1}{2} - \frac{1}{- 25} \Rightarrow \frac{1}{f} ≅ \frac{27}{50} \Rightarrow f ≅ 2 cm$

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(a) = x

For a normal eye, we have:
Far point at which the object can be placed, u = $\infty$
Distance between the eye lens and the retina, v = x
Thus, we have:
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{f} = \frac{1}{x} - \frac{1}{\infty} \Rightarrow f = x$ 1f=1v−1u⇒1f≅12−1−25⇒1f≅2750⇒f≅2 cm1f=1v−1u⇒1f≅12−1−25⇒1f≅2750⇒f≅2 cm

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(d) in each of these cases

The man is wearing glasses of positive power (converging lens). Hence, he cannot see nearby objects clearly. In other words, he is farsighted. Since he cannot see beyond 1 m, he is nearsighted. If a person with normal vision wears glasses of focal length +1 m, then the person will not be able to see beyond 1 m.

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(c) on f as well as u

The angular magnification is the ratio of the angle subtended by the image to the angle subtended by the object on an unaided eye.
In a simple microscope,
$m = \frac{\frac{h}{u}}{\frac{h}{D}} Here, u = Object distance from the lens D = Image distance from the lens h = Height of the object$

In normal adjustment, the object is placed at a distamce equal to focal length(f) from the lens and then magnification is given by m = D/f.
For u < f, m = D/f +1.

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(b) the power of the lens

For a simple microscope in normal adjustment, the object is placed at a distance equal to f (the focal length) from the lens, And the angular magnification is given by the relation
m = D/f.
For u < f, m = D/f + 1.
Power of the lens = 1/f
Angular magnification depends on power.

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(c) 1

$We have : h = Object height u = Object distance = 25 cm D = Near point = 25 cm Now, m = \frac{\frac{h}{u}}{\frac{h}{D}} \Rightarrow m = \frac{\frac{h}{25}}{\frac{h}{25}} \Rightarrow m = 1$

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(a) real
(d) inverted

The retina acts as a screen in the eye; only real, inverted images can be obtained on the screen.

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(a) Simple microscope
(d) Galilean telescope

In a simple microscope, the image formed is virtual and above the axis on the same side of the object. Similarly, in a Galilean telescope, the image formed is virtual, erect and magnified and between the objective lens and the eyepiece. A compound microscope and an astronomical telescope produce inverted images.

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(a) always strained in looking at an object

The maximum focal length of a normal eye is equal to the distance of the lens from the retina. In case it is greater than the distance, the eye will be strained while focusing the objects on the retina that is at a fixed distance from the eye lens.

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(c) Looking at a vertical tube containing some water

If the experimentalist is looking at a vertical tube containing some water, he has to be careful, as the lower meniscus will appear as upper.

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(a) If the far point goes ahead, the power of the divergent lens should be reduced.
(c) If the far point is 1 m away from the eye, the divergent lens should be used.

As the far point (x) is shifted ahead, the focal length (f) will be increased.
Thus, we have:
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{f} = \frac{1}{- x} - \frac{1}{- \infty} \Rightarrow \frac{1}{f} = \frac{1}{- x} \Rightarrow f = - x$

As power (P) is equal to the reciprocal of the focal length, it will be reduced. Also, because the focal length is negative, the lens used will be divergent when the far point is 1 m away.

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(b) L > u
(d) L > 2f₀

In a compound microscope, the objective lens of a short focal length, f_o, is used. The focal length of the objective lens is less than the focal length of the eyepiece, f_e, .The object is placed at a distance slightly greater than its focal length. The real, inverted image of the object forms somewhere in front of the eyepiece at a distance less than its focal length. This image acts as its object and the final image forms in between length, L, of the microscope. Please check and correct.
∴ L > f_o+ f_e> 2f_o
Also,
L > u

Page No 432:

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Given:
Distance of near point, D = 10 cm
Focal length, f = 10 cm
When the image is formed at the near point, D = 10 cm, the magnifying power $(m)$ of a simple microscope is given as:
$m = 1 + \frac{D}{f} = 1 + \frac{10}{10} = 1 + 1 = 2$
Thus, the maximum angular magnification is 2.

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Given:
Power of the objective lens = 25 D
The focal length of the objective lens is given by
$f_{0} = \frac{1}{25 D} = 0.04 m = 4 cm$
Power of the eyepiece = 5 D
The focal length of the eyepiece is given by
$f_{e} = \frac{1}{20 D} = 0.05 m = 5 cm$

(a) As f_o < fe, the instrument must be a microscope.

(b) Tube length, l = 25 cm
     Image distance for the eye lens, v_e = 25 cm
     We know:
     l = $v_{o} + f_{e}$
     $\Rightarrow v_{0} = l - f_{e}$ =25 $-$ 5 = 20 cm
$On applying the lens formula for the objective lens, we get : \frac{1}{v_{0}} - \frac{1}{u_{0}} = \frac{1}{f_{0}}$
$\Rightarrow - \frac{1}{u_{o}} = \frac{1}{4} - \frac{1}{20} \Rightarrow - \frac{1}{u_{0}} = \frac{5 - 1}{20} = \frac{4}{20} \Rightarrow u_{0} = \frac{20}{4} = 5 cm$
The angular magnification of microscope $(m)$ is given by
$m = \frac{v_{0}}{u_{0}} (\frac{D}{f_{e}}) = \frac{20}{5} (\frac{25}{5}) = 20 cm$
So, the required angular magnification of the microscope is 20.

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Let the tree make a visual angle θ with the eyes.
Thus, we have:
$θ = \frac{Height of the tree}{Distance from the eye}$
For tree A, $θ_{A} = \frac{2}{50} = 0.04$

For tree B, $θ_{B} = \frac{2.5}{80} = 0.03125$

For tree C, $θ_{C} = \frac{1.8}{70} = 0.0257$

For tree D, $θ_{D} = \frac{2.8}{100} = 0.028$
Now,
θ_A > θ_B > θ_D > θ_C
The more the value of θ, the bigger the apparent size of the tree.
So, the arrangement in decreasing order is A, B, D and C.

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For the simple microscope, we have:
Focal length, f = 12 cm
Least distance of clear vision, d = 25 cm
For maximum angular magnification,
Image distance = Least distance of clear vision
v = d = − 25 cm
$The lens formula is given by \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{u} = \frac{1}{v} - \frac{1}{f}$
On substituting the values, we get:
$\frac{1}{u} = \frac{1}{- 25} - \frac{1}{12} \frac{1}{u} = - \frac{25 + 12}{300} = - \frac{37}{300}$
$\Rightarrow u = - \frac{300}{37} = - 8.1 cm$
So, to produce maximum angular magnification, the object should be placed 8.1 cm away from the lens.

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For the simple microscope, we have:
Least distance of clear vision, D = 25 cm
Magnifying power, m = 3
Now,
Let the focal length be f.

(a) When the image is formed at D = 25 cm, the magnifying power of the simple microscope is given by
$m = 1 + (\frac{D}{f}) \Rightarrow 3 = 1 + (\frac{25}{f})$
$\Rightarrow \frac{25}{f} = 2 \Rightarrow f = \frac{25}{2} = 12.5 cm$

(b) When the image is formed at infinity in normal adjustment, the magnifying power is given by
$m = \frac{D}{f} = \frac{25}{12.5} = 2.0$

So, the magnifying power is 2 if the image is formed at infinity.

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For a simple microscope,
Magnification for a normal relaxed eye, m = 5
Least distant of distinct vision, D = 25 cm
Now,
Let the focal length be f.
The image will form at infinity, as the eye is relaxed.
The magnifying power $(m)$ of a simple microscope in normal adjustment is given by
$m = \frac{D}{f} \Rightarrow 5 = \frac{25}{f} \Rightarrow f = 5 cm$
For the relaxed far-sighted eye, D = 40 cm.
Magnification $(m)$ :
$m = \frac{D}{f} = \frac{40}{5} = 8$
The magnifying power for a simple microscope is 8X.

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For the compound microscope, we have:
Power of the objective lens = 25 D
The focal length of the objective lens is given by
$f_{0} = \frac{1}{25 D} = 0.04 m = 4 cm$
Power of the eyepiece = 5 D
The focal length of the eyepiece is given by
$f_{e} = \frac{1}{5 D} = 0.2 m = 20 cm$
Least distance of clear vision, D = 25 cm
Separation between the objective and the eyepiece, L = 30 cm
Magnifying power is maximum when the image is formed by the eyepiece at the least distance of clear vision, i.e., D = 25 cm.
For the eyepiece, we have:
v_e = − 25 cm and f_e = 20 cm
The lens formula is given by
$\frac{1}{v_{e}} = \frac{1}{u_{e}} + \frac{1}{f_{e}} \Rightarrow \frac{1}{u_{e}} = \frac{1}{v_{e}} - \frac{1}{f_{e}}$
$\Rightarrow \frac{1}{- 25} - \frac{1}{20} = - \frac{(4 + 5)}{100}$
$\Rightarrow u_{e} = \frac{- 100}{9} = 11.11 cm$
Let u_o and v_o be the object and image distance for the objective lens.
So, for the objective lens, the image distance will be
$V_{0} = L - u_{0}$
$v_{0} = 30 - 11.11 \Rightarrow v_{o} = 18.89 cm \dots (1)$
As the image produced is real, $v_{o} = + 18.89 cm$ .
The lens formula is given by
$\frac{1}{u_{o}} = \frac{1}{v_{o}} - \frac{1}{f_{o}} = \frac{1}{18.89} - \frac{1}{4} = - 0.197 u_{o} = - 5.07 cm \dots (2)$
Maximum magnifying power of the compound microscope:
$m = \frac{v_{0}}{u_{0}} (1 + \frac{D}{f_{e}}) = - (\frac{18.89}{- 5.07}) (1 + \frac{25}{20})$
= 3.7225 × 2.25 = 8.376
So, the maximum magnifying power of the compound microscope is 8.376.

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For the compound microscope, we have:
Focal length of the objective, f_o = 1.0 cm
Focal length of the eyepiece, f_e = 6 cm
Image distance from the eyepiece, v_e = $-$ 24 cm
Separation between the objective and the eyepiece = 9.8 cm to 11.8 cm
The lens formula is given by
$\frac{1}{v_{e}} - \frac{1}{u_{e}} = \frac{1}{f_{e}} \frac{1}{- 24} - \frac{1}{u_{e}} = \frac{1}{6}$
$\Rightarrow \frac{1}{u_{e}} = - \frac{1}{6} - \frac{1}{24} \Rightarrow \frac{1}{u_{e}} = - \frac{4 + 1}{24} = - \frac{5}{24} ∴ u_{e} = - \frac{24}{5} cm = - 4.8 cm$
(a) Separation between the objective and the eyepiece = 9.8 cm
So, for the objective lens, the image distance will be
v_o = 9.8 − 4.8 = 5 cm
The lens formula for the objective lens is given by
$\frac{1}{v_{0}} - \frac{1}{u_{0}} = \frac{1}{f_{0}} \frac{1}{5} - \frac{1}{u_{0}} = \frac{1}{1}$
$\Rightarrow - \frac{1}{u_{0}} = 1 - \frac{1}{5} = \frac{5 - 1}{5} \Rightarrow u_{0} = - \frac{5}{4} cm = - 1.25 cm$
Magnifying power of the compound microscope:
$m = \frac{v_{0}}{u_{0}} (1 + \frac{D}{f_{e}})$
$= \frac{5}{1.25} (1 + \frac{24}{6}) = 20$

(b) Separation between the objective and the eyepiece = 11.8 cm
We have:
m = 30
So, the required range of the magnifying power is 20–30.

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For the compound microscope, we have:
Power of the objective lens = 20 D
The focal length of the objective lens is given by
$f_{0} = \frac{1}{20 D} = 0.05 m = 5 cm$
Power of the eyepiece = 10 D
The focal length of the eyepiece is given by
$f_{e} = \frac{1}{10 D} = 0.1 m = 10 cm$
Least distance for clear vision, D = 25 cm,
To distinguish the two points having minimum separation, the magnifying power should be maximum.
For the eyepiece, we have:
Image distance, v_e = − 25 cm
Focal length, f_e = 10 cm
The lens formula is given by
$\frac{1}{u_{e}} = \frac{1}{v_{e}} - \frac{1}{f_{e}}$
$= \frac{1}{- 25} - \frac{1}{10} = \frac{- 2 - 5}{50} = \frac{- 7}{50}$
Separation between the objective and the eyepiece = 20 cm
So, the image distance for the objective lens $(v_{0})$ can be obtained as:
$v_{0} = 20 - \frac{50}{7} = \frac{90}{7} cm$
The lens formula for the objective lens is given by
$\frac{1}{u_{0}} = \frac{1}{v_{0}} - \frac{1}{f_{0}} = \frac{7}{90} - \frac{1}{5} = \frac{7 - 18}{90} = - \frac{11}{90}$
$\Rightarrow u_{0} = \frac{- 90}{11} cm$
Magnification of the compound microscope:
$m = \frac{v_{0}}{u_{0}} (1 + \frac{D}{f_{e}})$
$= - \frac{\frac{90}{7}}{\frac{- 90}{11}} (1 + \frac{25}{10}) = \frac{11}{7} \times (3.5) = 5.5$
∴ Minimum separation that the eye can distinguish = $\frac{0.22}{5.5}$ mm = 0.04 mm

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For the compound microscope, we have:
Magnifying power, m = 100
Focal length of the objective, f_o = 0.5 cm
Tube length, l = 6.5 cm
Since the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eyepiece.
∴ v_o + f_e = 6.5 cm …(1)
The magnifying power for normal adjustment is given by
$m = \frac{v_{o}}{u_{o}} \times \frac{D}{f_{e}} = - [1 - \frac{v_{o}}{f_{o}}] \frac{D}{f_{e}} \Rightarrow 100 = - [1 - \frac{v_{o}}{0.5}] \times \frac{25}{f_{e}} \Rightarrow 2 v_{o} - 4 f_{e} = 1 \dots (2)$
On solving equations (1) and (2), we get:
V_o = 4.5 cm and f_e = 2 cm
Thus, the focal length of the eyepiece is 2 cm.

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For the compound microscope, we have:
Focal length of the objective, f_o = 1.0 cm
Focal length of the eyepiece, f_e = 5 cm
Distance of the object from the objective, u₀ = 0.5 cm
Distance of the image from the eyepiece, v_e = 30 cm
The lens formula for the objective lens is given by
$\frac{1}{v_{0}} - \frac{1}{u_{0}} = \frac{1}{f_{0}} \Rightarrow \frac{1}{v_{0}} + \frac{1}{0.5} = \frac{1}{1}$
$\Rightarrow \frac{1}{v_{0}} = 1 - \frac{10}{5} = - 1 \Rightarrow v_{0} = - 1 cm$
The objective will form a virtual image at the side same as that of the object at a distance of 1 cm from the objective lens. The image formed by the objective will act as a virtual object for the eyepiece.
$The lens formula for the eyepiece is given by \frac{1}{v_{e}} - \frac{1}{u_{e}} = \frac{1}{f_{e}}$
$\Rightarrow \frac{1}{30} - \frac{1}{u_{e}} = \frac{1}{5} \Rightarrow - \frac{1}{u_{e}} = \frac{1}{5} - \frac{1}{30} = \frac{6 - 1}{30} = \frac{1}{6}$
⇒ u_e = − 6 cm
∴ Separation between the objective and the eyepiece = 6 − 1 = 5 cm

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For the astronomical telescope,
Magnifying power, m = 50
Length of the tube, L = 102 cm
Let the focal length of objective and eye piece be f₀ and f_e, respectively.
Now, using $m = \frac{f_{o}}{f_{e}}$ , we get:
f_o= 50f_e…(1)
And,
L = f_o + f_e = 102 cm …(2)
On substituting the value of f_o from (1) in (2), we get:
50fe + fe = 102
⇒ 51fe = 102
⇒ fe = 2 cm = 0.02 m
And,
f_o = 50 $\times$ 0.02 = 1 m
Power of the objective lens = $\frac{1}{f_{o}}$ = 1 D
And,
Power of the eye piece lens = $\frac{1}{f_{e}} = \frac{1}{0.02}$ = 50 D

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Answer:

In the astronomical telescope (in normal adjustment),
Focal length of the eyepiece, fe = 10 cm
Length of the tube, L = 1 m = 100 cm
Focal length of the objective, f_o = ?
We know:
f_o + f_e = L
$∴$ f_o = L – f_e = 100 – 10 = 90 cm
Magnifying power $(m)$ in normal adjustment:
m = $\frac{f_{o}}{f_{e}} = \frac{90}{10} = 9$

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The image will be formed at infinity.
Given:
Focal length of the objective, f_o = 30 cm
Length of the tube, L = 27 cm
Now,
L = $f_{o} - |f_{e}|$
In a Galilean telescope, the eyepiece lens is concave.
Thus, we have:
f_e = f_o – L = 30 – 27 = 3 cm

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Answer:

Here,
u = – 20 cm and v = – 50 cm
The lens formula is given by
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
On putting the values, we get:
$\frac{1}{f} = \frac{1}{(- 50)} - \frac{1}{(- 20)} = \frac{3}{100}$
$\Rightarrow f = \frac{100}{3} cm = \frac{1}{3} m$
∴ Power of the lens = $\frac{1}{f}$ = 3 D

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Answer:

For a near-sighted person, u = ∞.
Now, we have:
v = – 200 cm = – 2 m
The lens formula is given by
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
On putting the respective values, we get:
$\frac{1}{f} = \frac{1}{(- 2)} - \frac{1}{\infty} = - \frac{1}{2}$
∴ Power of the lens = $\frac{1}{f}$ = – 0.5 D

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The person must be near-sighted because the person wears glasses of power –2.5 D.
For a near-sighted person, u = ∞.
Focal length, f = $\frac{1}{Power}$ = $\frac{1}{(- 2.5)}$ = − 0.4 m = − 40 cm
The lens formula is given by
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Putting the values, we get:
$\frac{1}{v} = \frac{1}{\infty} + \frac{1}{(- 40)} \Rightarrow v = - 40$
So, the far point of the near-sighted person is 40 cm.

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Answer:

Given:
The professor can read the card at a distance of 25 cm using a glass of + 2.5 D.
After 10 years, he can read the farewell letter at a distance (u) of – 50 cm.
Focal length, f = $\frac{1}{Power}$ = $\frac{1}{2.5}$ = 0.4 m = 40 cm
The lens formula is given by
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Putting the values, we get:
$\frac{1}{v} = \frac{1}{(- 50)} + \frac{1}{40} \Rightarrow v = 200 cm$
∴ His near point, v = 200 cm
To read the farewell letter at a distance of 25 cm,
u = – 25 cm
The lens formula is given by
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Putting the values, we get:
$\frac{1}{f} = \frac{1}{(200)} - \frac{1}{(- 25)} = + \frac{9}{200} \Rightarrow f = \frac{200}{9} cm = \frac{2}{9} m$
∴ Power of the lens = $\frac{1}{f}$ = 4.5 D
He should use a lens of power + 4.5 D.

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Answer:

(a) When the eye lens is fully relaxed, we have:
u = ∞
Distance of the retina from the eye lens, v = 2 cm = 0.02 m
The lens formula is given by
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Putting the respective values, we get:
$\frac{1}{f} = \frac{1}{0.02} - \frac{1}{\infty} = \frac{1}{0.02}$
∴ Power of the lens = $\frac{1}{f}$ = 50 diopters
So, in a fully relaxed condition, the power of the eye lens is 50 D.

(b) When the eye lens is most strained:
u = – 25 cm = – 0.25 m
The lens formula is given by
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Putting the values, we get:
$\frac{1}{f} = \frac{1}{0.02} - \frac{1}{(- 0.25)} = 50 + 4 = 54$
∴ Power of the lens = $\frac{1}{f}$ = 54 diopters
So, in the most strained condition, the power of the eye lens is 54 D.

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Answer:

Given:
Near point of the child = 10 cm
Far point of the child = 100 cm
The retina is at a distance of 2 cm behind the eye lens.
Thus, we have:
v = 2 cm = 0.02 m
When the near point is 10 cm,
u = − 10 cm = − 0.1 m
v = 2 cm = 0.02 m
The lens formula is given by
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
On putting the respective values, we get:
$\frac{1}{f} = \frac{1}{0.02} - \frac{1}{(- 0.1)} = 50 + 10 = 60 m$
∴ Power of the lens, P = $\frac{1}{f}$ = 60 D
Now, consider the near point 100 cm.
Here,
u = − 100 cm = − 1 m
v = 2 cm = 0.02 m
The lens formula is given by
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Putting the values, we get:
$\frac{1}{f} = \frac{1}{0.02} - \frac{1}{(- 1)} = 50 + 1 = 51$
∴ Power of the lens = $\frac{1}{f}$ = 51 D
So, the range of the power of the eye lens is from + 60 D to + 51 D.

Page No 432:

Answer:

Given:
As the person wears spectacles at a distance of 1 cm from the eyes.
Distance of the image from the glass, v = Distance of the image from the eye − Separation between the glass and the eye
v = 25 − 1 = 24 cm = 0.24 m
For the near-sighted person, u = ∞.
The lens formula is given by
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
On substituting the values in the above formula, we get:
$\frac{1}{f} = \frac{1}{(- 0.24)} - \frac{1}{\infty} = - 42$
Power of the lens, P = $\frac{1}{f}$ = − 4.2 D
Thus, the power of the lens required to see distant objects is − 4.2 D.

Page No 432:

Answer:

Given: A person has near point of 100 cm.
According to the question, it is needed to read at a distance of 20 cm.
(a) When the contact lens is used:
u = – 20 cm = – 0.2 m,
v = – 100 cm = – 1 m
The lens formula is given by
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
On substituting the values in the above formula, we get:
$\frac{1}{f} = \frac{1}{(- 1)} - \frac{1}{(- 0.2)} = 4$
∴ Power of the lens, P = $\frac{1}{f}$ = 4 D
(b) When the person wears spectacles at a distance of 2 cm from the eyes:
u = – (20 – 2) = – 18 cm = – 0.18 m,
v = – 100 cm = –1 m
The lens formula is given by
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
On substituting the values, we get:
$\frac{1}{f} = \frac{1}{(- 1)} - \frac{1}{(- 0.18)} = 4.53$
∴ Power of the lens, P = $\frac{1}{f}$ = 4.53 D

Page No 432:

Answer:

Given:
The lady uses glasses of +1.5 D to have normal vision from 25 cm onwards.
Least distance of clear vision, D = 25 cm
Focal length of the glasses, f = $\frac{1}{power} = \frac{1}{1.5}$ m
She should have greater least distance of distinct vision without the glasses.
Take:
u = – 25 cm = $-$ 0.25 m
The lens formula is given by
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Putting the values, we get:
$\frac{1}{v} = 1.5 - \frac{1}{0.25} = 2.5 \Rightarrow v = 0.4 m = 40 cm$
Near point without glasses = 40 cm
Focal length of magnifying glass, f = $\frac{1}{20}$ = 0.05 m = 5 cm
(a) The maximum magnifying power $(m)$ if she uses the microscope together with her glass is given by
$m = 1 + \frac{D}{f}$
Here,
D = 25 cm
f = 5 cm
On substituting the values, we get:
$m = 1 + \frac{25}{5} = 6$
(b) Without the glasses, D = 40 cm.
$∴ m = 1 + \frac{D}{f} = 1 + \frac{40}{5} = 9$

Page No 432:

Answer:

Given:
For the left glass lens of the lady, we have:
v = – 40 cm and u = – 25 cm
The lens formula is given by
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Putting the values, we get:
$\frac{1}{f} = \frac{1}{(- 40)} - \frac{1}{(- 25)} = \frac{3}{200}$
$\Rightarrow f = \frac{200}{3} = 66.6 cm$

For the right glass lens of the lady, we have:
v = – 100 cm, u = – 25 cm
The lens formula is given by
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Putting the values, we get:
$\frac{1}{f} = \frac{1}{(- 100)} - \frac{1}{(- 25)} = \frac{3}{100}$
$\Rightarrow f = \frac{100}{3} = 33.3 cm$

(a) An astronomical telescope consists of two lenses: the objective lens having a large focal length and the eyepiece lens having a smaller focal length. So, she should use the right lens of focal length 33.3 cm as the eyepiece lens.

(b) With relaxed eye in normal adjustment,
f_o = 66.6 cm and f_e = 33.3 cm
Magnification $(m)$ in normal adjustment is given by
m = $\frac{f_{0}}{f_{e}}$
$∴ m = \frac{66.6 cm}{33.3 cm}$ = 2

So, with the relaxed eye, she can get the magnification of 2.

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