Question 16:

What is the difference between a real image and a virtual image? Give one example of each type of image.

Answer:

S. No.	Real Image	Virtual Image
1	An image that can be projected on a screen is called a real image.	An image that cannot be projected on a screen is called a virtual image.
2	The light rays actually meet at the point of image formation.	The light rays do not actually meet at the point of image formation, but appear to meet there.
3	Example: the image formed on the screen in a cinema hall.	Example: the image formed by a plane mirror.

S.No.	Wall Reflected Light	Mirror Reflected Light
1	Light rays are reflected in all directions due to the rough surface.	Light rays are reflected in one direction only due to the smooth surface.

(b) Regular reflection from smooth surfaces like mirrors leads to the formation of images.

Page No 174:

Question 22:

What is the difference between regular reflection of light and diffuse reflection of light? What type of reflection of light takes place from:

(a) a cinema screen
(b) a plane mirror
(c) a cardboard
(d) still water surface of a lake

Answer:

Regular Reflection	Diffused Reflection
In regular reflection, light is reflected in one direction only because of the smooth surface of the plane.	In diffused reflection, light is reflected in all directions because of the rough surface of the plane.

S.No.	Type of Surface	Type of Reflection
a	Cinema screen	Diffused reflection
b	Plane mirror	Regular reflection
c	Cardboard	Diffused reflection
d	Still water surface of a lake	Regular reflection

Plane mirrors are used to see ourselves. Example, mirrors in a bathroom.
Plane mirrors are fixed on the inner walls of certain shops ( like jewellery shops) to make them look bigger.
Plane mirrors are fitted at blind turns of some busy roads so that drivers can see the vehicles coming from the other side and prevent accidents.
Plane mirrors are used in periscopes.

Page No 174:

Question 26:

What is meant by 'reflection of light'? Define the following terms used in the study of reflection of light by drawing a labelled ray-diagram:

(a) Incident ray
(b) Point of incidence
(c) Normal
(d) Reflected ray
(e) Angle of incidence
(f) Angle of reflection

Answer:

The process of sending back light rays that fall on the surface of an object is called reflection of light. (Figure)

(a) Incident ray: The ray of light that falls on a mirror is called the incident ray. In the figure, AO is the incident ray of light. The incident ray gives the direction of the light falling on the mirror.

(b) Point of incidence: The point at which the incident ray falls on a mirror is called the point of incidence. In the figure, the point O on the surface of the mirror is the point of incidence.

(c) Normal: The normal is a line at right angles to the mirror surface at the point of incidence. In the figure, the line ON is the normal to the mirror surface at point O.

(d) Reflected ray: The ray of light that is sent back by the mirror is called the reflected ray. In the figure, OB is the reflected ray of light. The reflected ray of light shows the direction in which the light travels after being reflected from the mirror.

(e) Angle of incidence: The angle of incidence is the angle made by the incident ray with the normal at the point of incidence. In the figure, the angle AON is the angle of incidence. The angle of incidence is denoted by the letter i.

(f) Angle of reflection: The angle of reflection is the angle made by the reflected ray with the normal at the point of incidence. In the figure, the angle NOB is the angle of reflection. The angle of reflection is denoted by the letter r.

Page No 174:

Question 27:

State and explain the laws of reflection of light at a plane surface (like a plane mirror), With the help of a labelled ray-diagram. Mark the angles of 'incidence' and 'reflection' clearly on the diagram. If the angle of reflection is 47.5°, what will be the angle of incidence?

Answer:

a) The two laws of reflection are:

The angle of incidence is always equal to the angle of reflection . If the angle of incidence is i and the angle of reflection is r, then, according to the first law of reflection, $∠ i = ∠ r$ .

For example, if one measures the angle of reflection <NOB (in the figure), one will find that it is exactly equal to the angle of incidence <AON.

The second law of reflection states that the incident ray, the reflected ray and the normal (at the point of incidence), all lie in the same plane. For example, in the figure, the incident ray AO, the reflected ray OB and the normal ON, all lie in the same plane.

Concave mirror has a radius of curvature of 32 cm.
Therefore, R = -32 cm
Hence, the focal length of a concave mirror 'f' = $\frac{- 32}{2}$ = -16 cm

Page No 178:

Question 4:

If the focal length of a convex mirror is 25 cm, what is its radius of curvature?

Answer:

The focal length of a convex mirror is 25 cm, i.e. 'f' is 25 cm.

So, the radius of curvature of the convex mirror, 'R' = 2f = $2 \times 25$ = 50 cm

Page No 179:

Question 5:

Fill in the following blanks with suitable words:

(a) Parallel rays of light are reflected by a concave mirror to a point called the ..........
(b) The focal length of a concave mirror is the distance from the ......... to the mirror.
(c) A concave mirror .......... rays of light whereas convex mirror ............ rays of light
(d) For a convex mirror, parallel rays of light appear to diverge from a point called the ......... .

Answer:

(a) Parallel rays of light are reflected by a concave mirror to a point called the focus.
(b) The focal length of a concave mirror is the distance from the focus to the mirror.
(c) A concave mirror converges rays of light whereas convex mirror diverges rays of light.
(d) For a convex mirror, parallel rays of light appear to diverge from a point called the focus.

Page No 179:

Question 6:

What is a spherical mirror? Distinguish between a concave mirror and a convex mirror.

Answer:

A spherical mirror is a curved surface in which the inner or outer surface has the capability to reflect light.

Difference between convex and concave mirror are as follows:

Convex Mirror	Concave Mirror
1.In a convex mirror, reflection of light takes place on the bulged-out surface. 2. Convex mirror has virtual focus. 3. Convex mirror diverges the rays of light incident on it.	1.In a concave mirror, reflection of light takes place on the bent surface . 2. Concave mirror has real focus. 3. Concave mirror converges the rays of light incident on it.

(c) Pole (P) – Pole is the centre of the curved mirror surface, MM’.
(d) Principal axis – It is the line joining P and C.
(e) Aperture – It is the effective length of the mirror, i.e. MM’.

Page No 179:

Question 12:

(a) Define (i) principal focus of a concave mirror, and (ii) focal length of a concave mirror.
(b) Draw diagram to represent the action of a concave mirror on a beam of parallel light rays. Mark on this diagram principal axis, focus F, centre of curvature C, pole P and focal length f, of the concave mirror.

Answer:

(a) Definitions-

A diverging mirror is

(a) a plane mirror
(b) a convex mirror
(c) a concave mirror
(d) a shaving mirror

Answer:

(b) convex mirror

A convex mirror diverges the rays of light incident on it; hence, it acts as a diverging mirror.

Page No 179:

Question 16:

If R is the radius of curvature of a spherical mirror and f is its focal length, then:

(a) R = f
(b) R = 2f
(c) $R = \frac{f}{2}$
(d) R = 3f

Answer:

(b) R = 2f

If R is the radius of curvature of a spherical mirror and f is its focal length, then R = 2f.

Page No 179:

Question 17:

The focal length of a spherical mirror of radius of curvature 30 cm is:

(a) 10 cm
(b) 15 cm
(c) 20 cm
(d) 30 cm

Answer:

(b) 15 cm
The focal length of a spherical mirror is half of its radius of curvature.

Page No 179:

Question 18:

If the focal length of a spherical mirror is 12.5 less cm, its radius of curvature will be:

(a) 25 cm
(b) 15 cm
(c) 20 cm
(d) 35 cm

Answer:

(a) 25 cm

Radius of curvature 'R' = 2f

Using the following rules for image formation:

A ray of light passing through the centre of curvature of a concave mirror is reflected along the same path, and
A ray of light passing through the focus of a concave mirror becomes parallel to the principal axis after reflection,

we get:

Page No 189:

Question 9:

Draw the following diagram in your answer book and show the formation of image with the help of suitable rays:
Figure

Answer:

Using the following rules for image formation:

A ray of light that is parallel to the principal axis of a concave mirror passes through its focus after reflection, and
A ray of light passing through the focus of a concave mirror becomes parallel to the principal axis after reflection,

we get:

between the focus and centre of curvature
real and inverted
smaller than the object (or diminished)

Page No 189:

Question 16:

If an object is placed at a distance of 8 cm from a concave mirror of focal length 10 cm, discuss the nature of the image formed by drawing the ray diagram.

Answer:

As shown in the diagram, the image formed is

behind the mirror
virtual and erect
larger than the object

Shaving mirrors
By dentists, to see the teeth of patients

The reason being, an enlarged image of an object is seen when it is held within the focus of a concave mirror.

Page No 190:

Question 27:

(a) Draw ray-diagrams to show the formation of images when the object is placed in front of a concave mirror (converging mirror):
(i) between its pole and focus
(ii) between its centre of curvature and focus
Describe the nature, size and position of the image formed in each case.
(b) State one use of concave mirror based on the formation of image as in case (i) above.

Answer:

(i) When an object is placed between the pole and the focus of a concave mirror, the image formed is virtual, erect, magnified and behind the mirror. This is illustrated as follows:

Here, the image A'B' of the object AB, placed between the pole and the focus of a concave mirror, is virtual and erect.

(ii) When an object is placed between the centre of curvature and the focus of a concave mirror, the image formed is real, inverted, and enlarged. This is illustrated as follows:

Here, the image A'B' of the object AB, placed between the centre of curvature and the focus of a concave mirror, is real and inverted.

(iii) The concave mirror can be used in the following, on the bases of image formation as demonstrated in case (i):

As a shaving mirror
As a make-up mirror
As a dentist's mirror

Page No 190:

Question 28:

(a) Give two circumstances in which a concave mirror can form a magnified image of an object placed in front of it. Illustrate your answer by drawing labelled ray diagrams for both.
(b) Which one of these circumstances enables a concave mirror to be used as a shaving mirror?

Answer:

(a) The two circumstances in which a concave mirror can form a magnified image of an object that is placed in front of it are:

When an object is placed between the pole and the focus of a concave mirror, the image formed is virtual, erect and magnified, as shown in the figure. Here, the object AB is placed between the pole and the focus of a concave mirror. Its image A'B' is formed behind the mirror and is virtual, erect and magnified.
When an object is placed between the centre of curvature and the focus of a concave mirror, the image formed is real, inverted and enlarged, as shown in figure 2. Here, the object AB is placed between the centre of curvature and the focus of a concave mirror. Its image A'B' is formed beyond the centre of curvature and is real, inverted and enlarged.

According to the new cartesian sign convention :

All distances are measured from the pole of the mirror as origin.
Distances measured in the same direction as that of incident light are taken as positive.
Distances measured against the direction of incident light are taken as negative.
Distances measured upward and perpendicular to the principal axis are taken as positive.
Distances measured downward and perpendicular to the principal axis are taken as negative.

Page No 193:

Question 8:

Giving reasons, state the 'signs' (positive or negative) which can be given to the following:

(a) object distance (u) for a concave mirror or convex mirror
(b) image distance (v) for a concave mirror
(c) image distance (v) for a convex mirror

Answer:

(a) Object distance (u) for a concave or convex mirror is given a negative sign. This is because, the object is always placed to the left of the mirror.
(b)

The image distance (v) for a concave mirror is given a positive sign if the image formed is behind the mirror and to the right of it.
The image distance (v) for a concave mirror is given a negative sign if the image formed is in front of the mirror and to the left of it.

(a) If the magnification has a plus sign, then the image is virtual and erect.
(b) If the magnification has a minus sign, then the image is real and inverted.

Page No 198:

Question 11:

An object is placed at a distance of 10 cm from a concave mirror of focal length 20 cm.

(a) Draw a ray diagram for the formation of image.
(b) Calculate the image distance.
(c) State two characteristics of the image formed.

Answer:

(a) Ray Diagram-

(b) Image Distance-

Distance of the object from the mirror 'u' = −10 cm
Focal length of the concave mirror 'f' = −20 cm

We have to find the distance of the image 'v'.

Mirror Formula:

\frac{1}{f} = \frac{1}{v} + \frac{1}{u}

Therefore

\frac{1}{- 20} = \frac{1}{v} + \frac{1}{- 10} \frac{1}{v} = - \frac{1}{20} + \frac{1}{10} = - \frac{1}{20} + \frac{2}{20} \frac{1}{v} = \frac{1}{20} v = 20 cm

Thus, the distance of the image 'v' is 20 cm.

(1) It is virtual and erect.

(2) It is magnified.

Page No 198:

Question 12:

If an object of 10 cm height is placed at a distance of 36 cm from a concave mirror of focal length 12 cm, find the position, nature and height of the image.

Answer:

Distance of the object from the mirror 'u' = −36 cm
Height of the object 'h_o' = 10 cm
Focal length of the mirror 'f' = −12 cm
We have to find the distance of the image 'v'.
Height of the image h_i = ?
Using the mirror formula, we get
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \frac{1}{- 12} = \frac{1}{v} + \frac{1}{- 36} \frac{1}{v} = \frac{1}{36} - \frac{1}{12} \frac{1}{v} = \frac{1 - 3}{36} = \frac{- 2}{36} v = - 18 cm$

Distance of the image 'v' = −18 cm
Using the magnification formula, we get
$m = \frac{h_{i}}{h_{o}} = - \frac{v}{u} \frac{h_{i}}{10} = \frac{- (- 18)}{- 36} = - \frac{1}{2} \frac{h_{i}}{10} = - \frac{1}{2} h_{i} = - 5 cm$

Height of the image 'h_i' = −5 cm
Again using the magnification formula, we get
$m = \frac{- v}{u} = \frac{- (- 18)}{36} m = - \frac{1}{2}$

Thus, the image is real, inverted and small in size.

Page No 198:

Question 13:

At what distance from a concave mirror focal length 10 cm should an object 2 cm long be placed in order to get an erect image 6 cm tall?

Answer:

Height of the object 'h_o' = 2 cm

Focal length of the mirror 'f' = −10 cm
Height of the image 'h_i' = 6 cm
We have to find the distance of the object from the mirror 'u'.
Using the magnification formula, we get
$m = \frac{h_{i}}{h_{o}} = - \frac{v}{u} m = \frac{6}{2} = - \frac{v}{u} m = 3 = - \frac{v}{u} thus v = - 3 u$

Now, using the mirror formula, we get

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \frac{1}{- 10} = \frac{1}{- 3 u} + \frac{1}{u} \frac{1}{10} = \frac{1}{3 u} - \frac{3}{3 u} = \frac{- 2}{3 u} u = \frac{- 2 \times 10}{3} u = - 6.67 cm$

Thus, the distance of the object from the mirror 'u' is −6.67 cm.

Page No 198:

Question 14:

When an object is placed at a distance of 15 cm from a concave mirror, its image is formed at 10 cm in front of the mirror. Calculate the focal length of the mirror.

Answer:

Distance of the object from the mirror 'u' = −15 cm

Distance of the image 'v' = −10 cm
We have to find the focal length of the mirror 'f'.
Using the mirror formula, we get
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \frac{1}{f} = \frac{1}{- 10} + \frac{1}{- 15} = - \frac{1}{10} - \frac{1}{15} \frac{1}{f} = - \frac{3}{30} - \frac{2}{30} \frac{1}{f} = - \frac{5}{30} f = - 6 cm$

Thus, the focal length of the concave mirror 'f' is 6 cm.

Page No 198:

Question 15:

An object 3 cm high is placed at a distance of 8 cm from a concave mirror which produces a virtual image 4.5 cm high:

(i) What is the focal length of the mirror?
(ii) What is the position of image?
(iii) Draw a ray-diagram to show the formation of image.

Answer:

Distance of the object from the mirror 'u' = -8 cm

Height of the object 'h_o' = 3 cm
Height of the image 'h_i' = 4.5 cm
We have to find the focal length of the mirror 'f' and distance of the image 'v'.
Using the magnification formula, we get
$m = \frac{h_{i}}{h_{0}} = - \frac{v}{u} m = \frac{4.5}{3} = \frac{- v}{- 8} v = \frac{4.5 \times 8}{3} = 12 cm$

Therefore, the distance of the image 'v' is 12 cm behind the mirror.

Now, using the mirror formula, we get

\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{12} + \frac{1}{- 8} \frac{1}{f} = \frac{2}{24} - \frac{3}{24} = \frac{- 1}{24} f = - 24 cm

Thus, the focal length of the concave mirror 'f' is 24 cm.

Page No 198:

Question 16:

A converging mirror forms a real image of height 4 cm of an object of height 1 cm placed 20 cm away from the mirror:

(i) Calculate the image distance.
(ii) What is the focal length of the mirror?

Answer:

A concave mirror is a converging mirror.

Distance of the object from the mirror 'u' = -20 cm

Height of the object 'h_o' = 1 cm

Height of the image 'h_i' = -4 cm
We have to find the distance of the image 'v' and focal length of the mirror 'f'.

Using the magnification formula, we get

m = \frac{h_{i}}{h_{o}} = - \frac{v}{u} \frac{- 4}{1} = \frac{- v}{- 20} v = \frac{20 \times - 4}{1} = - 80 cm

Thus, the distance of the image 'v' is 80 cm.

Now, using the mirror formula, we get

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \frac{1}{f} = \frac{1}{- 80} + \frac{1}{- 20} = - \frac{1}{80} - \frac{4}{80} = - \frac{5}{80} \frac{1}{f} = - \frac{1}{16} f = - 16 cm$

Thus, the focal length of the concave mirror is 16 cm.

Page No 198:

Question 17:

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained? Find the size and nature of image.

Answer:

Distance of the object from the mirror 'u' = −27 cm
Height of the object 'h_o' = 7 cm
Focal length of the mirror 'f' = −18 cm
We have to find the distance of the image 'v' and height of the image 'h_i'.
Using the mirror formula, we get
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

$\Rightarrow \frac{1}{- 18} = \frac{1}{v} + \frac{1}{- 27} \Rightarrow \frac{1}{- 18} = \frac{1}{v} - \frac{1}{27} \Rightarrow \frac{1}{27} - \frac{1}{18} = \frac{1}{v} \Rightarrow \frac{- 1}{54} = \frac{1}{v} \Rightarrow v = - 54 cm$

Thus, the distance of the image 'v' is 54 cm.
Now, using the magnification formula, we get
$m = \frac{h_{i}}{h_{o}} = - \frac{v}{u} \frac{h_{i}}{7} = - \frac{(- 54)}{(- 27)} = - \frac{2}{1} h_{i} = - 14 cm$

Therefore, the height of the image 'h_i' is 14 cm.
Again, using the magnification formula, we get
$m = - \frac{v}{u} = - \frac{(- 54)}{(- 27)} m = - 2$

Thus, the image is real, inverted and large in size.

Page No 199:

Question 18:

An object 3 cm high is placed at a distance of 10 cm in front of a converging mirror of focal length 20 cm. Find the position, nature and size of the image formed.

Answer:

Distance of the object from the mirror 'u' = −10 cm
Height of the object 'h_o' = 3 cm
Focal length of the mirror 'f' = −20 cm
We have to find the distance of the image 'v' and the height of the image 'h_i'.
Using the mirror formula, we get
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

$\frac{1}{- 20} = \frac{1}{v} + \frac{1}{- 10} \Rightarrow \frac{1}{v} = \frac{1}{- 20} + \frac{1}{10} \Rightarrow \frac{1}{v} = \frac{1}{20} \Rightarrow v = 20 cm$
Thus, the distance of the image 'v' is 20 cm other side of cancave mirror.
Now, using the magnification formula, we get
$m = \frac{h_{i}}{h_{o}} = \frac{- v}{u}$
$\frac{h_{i}}{h_{o}} = \frac{- 20}{10}$
$\frac{h_{i}}{3} = - 2$
h_i= −6 cm
Therefore, the height of the image 'h_i' is 6 cm.
Again, using the magnification formula, we get
$m = \frac{- v}{u}$

$m = \frac{- 20}{10} m = - 2 cm$

Thus, the image is virtual, erect.

Page No 199:

Question 19:

A concave mirror has a focal length of 4 cm and an object 2 cm tall is placed 9 cm away from it. Find the nature, position and size of the image formed.

Answer:

Distance of the object from the mirror 'u' = −9cm
Height of the object 'h_o' = 2 cm
Focal length of the mirror 'f' = −4 cm
We have to find the distance of the image 'v' and height of the image 'h_i'.
Using the mirror formula, we get
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \frac{1}{- 4} = \frac{1}{v} + \frac{1}{- 9} \frac{1}{- 4} = \frac{1}{v} - \frac{1}{9} \frac{1}{9} - \frac{1}{4} = \frac{1}{v} \frac{4 - 9}{36} = \frac{1}{v} \frac{- 5}{36} = \frac{1}{v} = \frac{- 1}{7.2} v = 7.2 cm$

Therefore, distance of the image 'v' is 7.2 cm.
Now, using the magnification formula, we get
$m = \frac{h_{i}}{h_{o}} = - \frac{v}{u} \frac{h_{i}}{h_{o}} = - \frac{(- 7.2)}{- 9} \frac{h_{i}}{2} = - \frac{(- 7.2)}{- 9} h_{i} = \frac{2 \times (- 7.2)}{9} h_{i} = - 1.6 cm$

Thus, the height of the image 'h_i' is 1.6 cm.
Again, using the magnification formula, we get
$m = - \frac{v}{u} m = \frac{- (- 7.2)}{- 9} m = - 0.8$

Thus, the image is real, inverted and small in size.

Page No 199:

Question 20:

When an object is placed 20 cm from a concave mirror, a real image magnified three times is formed. Find:

(a) the focal length of the mirror.
(b) Where must the object be placed to give a virtual images three times the height of the object?

Answer:

Given,

Distance of the object 'u' = -20 cm
Magnification 'm' = −3
(a)We have to find the focal length of the mirror.

Using the magnification formula, we get

m = - \frac{v}{u}

- 3 = \frac{- v}{- 20}

v = −60 cm

Therefore, the distance of the image 'v' is −60 cm.
Using the mirror formula, we get

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

$\frac{1}{f} = \frac{1}{- 60} + \frac{1}{- 20}$

$\frac{1}{f} = - \frac{1}{60} - \frac{1}{20}$

$\frac{1}{f} = - \frac{1}{60} - \frac{3}{60} = - \frac{4}{60}$
$\frac{1}{f} = \frac{- 1}{15}$

f = −15 cm

Thus, the focal length of the concave mirror is 15 cm.

(b) Now, if the image is virtual and 3 times magnified, then we have to find the position of the object.

Given,

Focal length of the concave mirror = −15 cm

Magnification 'm' = 3

Using the magnification formula, we get

m = - \frac{v}{u}

3 = - \frac{v}{u}

$v = - 3 u$

Therefore, the distance of the image 'v' is −3u

Again, using the mirror formula, we get

\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \frac{1}{f} = \frac{1}{- 3 u} + \frac{3}{3 u} = \frac{2}{3 u} f = \frac{3 u}{2} u = \frac{2 f}{3} = \frac{2 \times (- 15)}{3} u = - 10 cm

Thus, the object should be placed at a distance of 10 cm to get a virtual image three times the height of the object.

Page No 199:

Question 21:

A dentist's mirror has a radius of curvature of 3 cm. How far must it be placed from a small dental cavity to give a virtual image of the cavity that is magnified five times?

Answer:

Given,

The dentist's mirror is a concave mirror.

Radius of curvature 'R' = -3 cm

Therefore, the

focal length of the mirror f = \frac{R}{2} = \frac{- 3}{2} = - 1.5 cm

Given that magnification 'm' = 5

Therefore, using the magnification formula, we get

m = - \frac{v}{u} v = - mu

v=-5u
Now, using the mirror formula, we get

\frac{1}{f} = \frac{1}{v} + \frac{1}{u}

\frac{1}{- 1.5} = \frac{1}{- 5 u} + \frac{1}{u} \Rightarrow \frac{1}{- 1.5} = \frac{1}{- 5 u} + \frac{1}{u} \Rightarrow \frac{- 1}{1.5} = \frac{1}{- 5 u} + \frac{5}{5 u} = \frac{4}{5 u} \Rightarrow u = \frac{(- 1.5) \times 4}{5} = \frac{- 6}{5} \Rightarrow u = - 1.2 cm

The dentist should place the object at a distanc of 1.2 cm from the mirror.

Page No 199:

Question 22:

A large concave mirror has a radius of curvature of 1.5 m. A person stands 10 m in front of the mirror. Where is the person's image?

Answer:

Given,
The radius of curvature of the mirror 'R' = -1.5m
Focal length of the mirror $' f' = \frac{1}{R} = \frac{1}{- 1.5} = - 0.75 m$
Distance of the object 'u' =-10 m
We have to find the distance of the image 'v'.
Using the mirror formula, we get

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

$\frac{1}{0.75} = \frac{1}{v} + \frac{1}{- 10} - \frac{100}{75} = \frac{1}{v} - \frac{1}{10} or \frac{1}{v} = \frac{1}{10} - \frac{100}{75} or \frac{1}{v} = \frac{1}{10} - \frac{4}{3} or \frac{1}{v} = \frac{3}{30} - \frac{40}{30} or \frac{1}{v} = \frac{- 37}{30} or v = \frac{- 30}{37} or v = 0.81 m$
Thus, the person's image will be formed at a distance of 0.81 m from the mirror.

Page No 199:

Question 23:

An object of 5.0 cm size is placed at a distance of 20.0 cm from a converging mirror of focal length 15.0 cm. At what distance from the mirror should a screen be placed to get the sharp image? Also calculate the size of the image.

Answer:

Concave mirror is a converging mirror.
Distance of the object from the mirror 'u' = -20 cm

Height of the object 'h_o' = 5 cm

Focal length of the mirror 'f' = -15 cm

We have to find the distance of the image 'v' and the height of the image 'h_i'.

Using the mirror formula, we get

\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \frac{1}{- 15} = \frac{1}{v} + \frac{1}{- 20} = \frac{1}{v} - \frac{1}{20} or \frac{1}{v} = \frac{1}{20} - \frac{1}{15} or \frac{1}{v} = \frac{3 - 4}{60} = \frac{1}{60} or v = - 60 cm Thus, the image is at a distance of 60 cm in front of the mirror .

$Now, using the magnification formula, we get m = \frac{- v}{u} = \frac{h_{i}}{h_{o}} or \frac{h_{i}}{5} = \frac{- (- 60)}{- 20} or \frac{h_{i}}{5} = - 3 or h_{i} = - 3 \times 5 = - 15 cm Therefore, the height of the image will be 15 cm, and the negative sign shows that the image will be formed below the principal axis .$

Page No 199:

Question 24:

A concave mirror produces three times enlarged virtual image of an object placed at 10 cm in front of it. Calculate the radius of curvature of the mirror.

Answer:

Given,
Distance of the object 'u' = -10 cm
Magnification 'm' = 3
We have to find the focal length 'f' and the radius of curvature 'R'.
Using the magnification formula, we get
$m = \frac{- v}{u} 3 = \frac{- v}{- 10} v = 30 cm Thus, the distance of the image' v' is 30 cm behind the mirror . N o w, u \sin g t h e mirror formula, w e g e t \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \frac{1}{f} = \frac{1}{30} + \frac{1}{- 10} or \frac{1}{f} = \frac{1}{30} - \frac{3}{30} = \frac{- 2}{30} or \frac{1}{f} = - \frac{1}{15} or f = - 15 cm Thus, the focal length of the mirror' f' is - 15 cm . Therefore, the radius of curvature of the mirror' R' will be 2 f = 2 \times - 15 = - 30 cm Radius of curvature of the mirror is 30 cm .$

Page No 199:

Question 25:

A bright object 50 mm high stands on the axis of a concave mirror of focal length 100 mm and at a distance of 300 mm from the concave mirror. How big will the image be?

Answer:

Distance of the object from the mirror 'u' = -300 mm

Height of the object 'h_o' = 50 mm

Focal length of the mirror 'f' = -100 mm

We have to find the distance of the image 'v' and the height of the image 'h_i'.

\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \frac{1}{- 100} = \frac{1}{v} + \frac{1}{- 300} - \frac{1}{100} = \frac{1}{v} - \frac{1}{300} \frac{1}{v} = \frac{1}{300} - \frac{1}{100} \frac{1}{v} = \frac{1}{300} - \frac{1}{100} \frac{1}{v} = \frac{- 2}{300} v = - 150 mm Thus, the image will be at a distance of 150 mm in front of the mirror . Now, using the magnification formula, we get m = \frac{h_{i}}{h_{o}} = \frac{- v}{u} \frac{h_{i}}{50} = \frac{- (- 150)}{- 300} = - \frac{1}{2} h_{i} = - \frac{50}{2} = - 25 mm Thus, the size of the image will be 25 mm .

Page No 199:

Question 26:

How far should an object be placed from the pole of a converging mirror of focal length 20cm to form a real image of the size exactly $\frac{1}{4}$ th the size of the object?

Answer:

Given,
Focal length of the concave mirror (f) = 20 cm
Magnification (m) = -1/4
Using the magnification formula, we get

$m = \frac{- v}{u} \frac{- 1}{4} = \frac{- v}{u} v = \frac{u}{4} U \sin g t h e mirror formula, w e g e t \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \frac{1}{- 20} = \frac{4}{u} + \frac{1}{u} \frac{- 1}{20} = \frac{5}{u} u = - 100 cm Thus, the object should be placed in front of t h e c o ncave mirror at a distance of 100 cm .$

Page No 199:

Question 27:

When an object is placed at a distance of 50 cm from a concave spherical mirror, the magnification produced is, $- \frac{1}{2}$ . Where should the object be placed to get a magnification of, $- \frac{1}{5}$ ?

Answer:

Given,
Magnification $(m) = \frac{- 1}{2}$

Distance of the object (u) = −50 cm

We have to find the distance of the object when the magnification is

m = - \frac{1}{5}

.
Therefore, using the magnification formula, we get

$m = \frac{- v}{u}$

v = −mu
$v = - (- \frac{1}{2} \times - 50)$
$v = - 25 cm$

Now, using the mirror formula, we get

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

$\frac{1}{f} = \frac{1}{- m u} + \frac{1}{u}$

$\frac{1}{f} = \frac{1}{- 25} + \frac{1}{- 50}$

$o r \frac{1}{f} = - \frac{2}{50} - \frac{1}{50} = - \frac{3}{50} or f = - \frac{50}{3} cm$
Now,
Magnification = $m = - \frac{1}{5}$

Therefore, using the mirror formula, we get
$\frac{1}{f} = \frac{1}{- m u} + \frac{1}{u} or \frac{- 3}{50} = \frac{- 5}{- 5} + \frac{1}{u} or \frac{- 3}{50} = \frac{5}{u} + \frac{1}{u} or \frac{- 3}{50} = \frac{6}{u} u = - \frac{50 \times 6}{3} = - 100 cm The object should be placed in front of the mirror at a distance of 100 cm to get the magnefiction of - \frac{1}{5} .$

Page No 199:

Question 28:

An object is placed (a) 20 cm, (b) 4 cm, in front of a concave mirror of focal length 12 cm. Find the nature and position of the image formed in each case.

Answer:

Case (a)
Distance of the object from the mirror (u) = -20 cm
The focal length of the concave mirror (f) = -12 cm
We have to find the distance of the image (v).
Using the magnification formula, we get
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} So or \frac{1}{- 12} = \frac{1}{v} + \frac{1}{- 20} \frac{1}{- 12} = \frac{1}{v} - \frac{1}{20} o r \frac{1}{20} - \frac{1}{12} = \frac{1}{v} o r \frac{3 - 5}{60} = \frac{1}{v} o r \frac{- 2}{60} = \frac{1}{v} o r v = \frac{60}{- 2} v = - 30 cm T he image will be at a distace of 30 cm in front of mirror . A g a i n, M agn i fication (m) = \frac{- v}{u} m = \frac{- (- 30)}{- 20} = - 1.5 T h e r e f o r e, t he image will be real, inverted and enlarged .$

Thus the image is real, inverted and enlarged.
Case B
Distance of the object from the mirror (u) = -4 cm
The focal length of the concave mirror (f) = -12 cm
We have to find the distance of the image (v).
Using the mirror formula, we get
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \frac{1}{- 12} = \frac{1}{v} + \frac{1}{- 4} or \frac{1}{4} - \frac{1}{12} = \frac{1}{v} or \frac{3 - 1}{12} = \frac{1}{v} or \frac{2}{12} = \frac{1}{v} v = 6 cm T h e r e f o r e, d istance of image (v) is 6 cm . N o w, u \sin g t h e magnif i c a t i o n formula, w e g e t m = \frac{- v}{u} m = \frac{- 6}{- 4} = 1.5 T h u s, the image is virtual and er e ct .$

Page No 199:

Question 29:

A concave mirror produces a real image 1 cm tall of an object 2.5 mm tall placed 5 cm from the mirror. Find the position of the image and the focal length of the mirror.

Answer:

Distance of the object from the mirror (u) = -5 cm
Height of the image (h_i) = -1 cm
Height of the object (h_o) = 2.5 mm = 0.25 cm
We have to find the image distance (v) and the focal length of the mirror (f).
Using the magnification formula, we get
$m = \frac{h_{i}}{h_{o}} = \frac{- v}{u} or \frac{- 1}{0.25} = \frac{- v}{- 5} or v = \frac{- 5}{0.25} = - 20 cm$
Thus, the distance of the image is -20 cm.
It means that the image will form 20 cm in front of the mirror.
Now, using the mirror formula, we get
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

$\frac{1}{f} = \frac{1}{- 20} + \frac{1}{- 5} or \frac{1}{f} = - \frac{1}{20} - \frac{1}{5} or \frac{1}{f} = - \frac{1}{20} - \frac{4}{20} = \frac{- 5}{20} = \frac{- 1}{4} f = - 4 cm$

Thus, the focal length of the mirror is 4 cm.

Page No 199:

Question 30:

A man holds a spherical shaving mirror of radius of curvature 60 cm, and focal length 30 cm, at a distance of 15 cm, from his nose. Find the position of image, and calculate the magnification.

Answer:

Distance of the object, i.e. nose, from the mirror (u) = -15 cm
The focal length of the concave mirror (f) = -30 cm

We have to find the distance of the image (v).

Using the mirror formula, we get

\frac{1}{f} = \frac{1}{v} + \frac{1}{u}

\frac{1}{- 30} = \frac{1}{v} - \frac{1}{15}

or \frac{1}{v} = \frac{1}{15} - \frac{1}{30} or \frac{1}{v} = \frac{2 - 1}{30} or \frac{1}{v} = \frac{1}{30} v = 30 cm

Thus, the position of the image is 30 cm behind the mirror.

Now, using the magnification formula, we get

m = \frac{- v}{u}

m = \frac{- 30}{- 15} = 2

Therefore, the image is 2 times magnified than the object.

Page No 199:

Question 31:

(a) An object is placed just outside the principal focus of concave mirror. Draw a ray diagram to show how the image is formed, and describe its size, position and nature.
(b) If the object is moved further away from the mirror, what changes are there in the position and size of the image?
(c) An object is 24 cm away from a concave mirror and its image is 16 cm from the mirror. Find the focal length and radius of curvature of the mirror, and the magnification of the image.

Answer:

(a) Ray Diagram-

(b) The image will move towards the mirror, and its size will gradually decrease.
(c) Distance of object (u) = -24 cm
Distance of image (v) = -16 cm
We have to find the focal length of the mirror (f) and the magnification of the image (m).
Using the mirror formula, we get

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

\frac{1}{f} = \frac{1}{- 24} + \frac{1}{- 16} or \frac{1}{f} = \frac{- 2 - 3}{48} o r \frac{1}{f} = \frac{- 5}{48} o r f = \frac{- 48}{5} f = - 9.6 cm T h e r e f o r e, t h e focal l e n g t h of t h e concave mirror is 9.6 cm . R adius of curvature (R) = 2 f R = 2 \times - 9.6 R = - 19.2 cm T h e r e f o r e, t h e radius of curvature of the mirror is 19.2 cm

Now, using the magnification formula, we get

m = \frac{- v}{u}

m = \frac{- 16}{- 24} m = 0.66

Thus, the image is real, inverted and small in size.

Page No 199:

Question 32:

Linear magnification produced by a concave mirror may be:

(a) less than 1 or equal to 1
(b) more than 1 or equal to 1
(c) less than 1, more than 1 or equal to 1
(d) less than 1 or more than 1

A concave mirror produces a magnification of +4 when the object is placed between the focus and the pole.

Page No 200:

Question 37:

If a magnification of, −1 (minus one) is to be obtained by using a converging mirror, then the object has to be placed:

(a) between pole and focus
(b) at the centre of curvature
(c) beyond the centre of curvature
(d) at infinity

Answer:

(b) at the centre of curvature

If a magnification of −1 (minus one) is to be obtained by using a converging mirror, the object needs to be placed at the centre of curvature so that an image of same size as the object can be formed.

Page No 200:

Question 38:

In order to obtain a magnification of, −0.6 (minus 0.6) with a concave mirror, the object must be placed:

(a) at the focus
(b) between pole and focus
(c) between focus and centre of curvature
(d) beyond the centre of curvature

Answer:

Answer:

(b) is less than one

Linear magnification (m) produced by a rear view mirror, installed in vehicles, is less than one.

Page No 200:

Question 42:

Between which two points of concave mirror should an object be placed to obtain a magnification of:

(a) −3
(b) +2.5
(c) −0.4

Answer:

(a) An object, in front of a concave mirror, should be placed between the latter's centre of curvature (C) and focus (F) to obtain a magnification of −3.
(b) An object, in front of a concave mirror, should be placed between the latter's focus (F) and the pole (P) to obtain a magnification of +25.
(c) An object, in front of a concave mirror, should be placed beyond the latter's centre of curvature (C) to obtain a magnification of −0.4.

Page No 200:

Question 43:

At what distance from a concave mirror of focal length 10 cm should an object be placed so that:

(a) its real image is formed 20 cm from the mirror?
(b) its virtual image is formed 20 cm from the mirror?

Answer:

(a)

G i v e n, I t i s a c o n c a v e m i r r o r . f = - 10 cm v = - 20 cm S ince t h e image is real a n d form s in front of the mirror, t h e e q u a t i o n w i l l b e \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \Rightarrow \frac{1}{- 10} = \frac{1}{- 20} + \frac{1}{u} \Rightarrow \frac{1}{u} = \frac{1}{- 10} + \frac{1}{20} \Rightarrow \frac{1}{u} = \frac{- 20 + 10}{200} \Rightarrow \frac{1}{u} = \frac{- 10}{200} \Rightarrow \frac{1}{u} = - \frac{1}{20} \Rightarrow u = - 20 cm T h e r e f o r e, t he object should be placed at a distance of 20 cm f r o m t h e m i r r o r to form a real image .

(b)

G i v e n, f = - 10 cm v = + 20 cm S ince t h e image is virtual a n d form s behind the mirror, t h e e q u a t i o n w i l l b e \Rightarrow \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \Rightarrow \frac{1}{- 10} = \frac{1}{20} + \frac{1}{u} \Rightarrow - \frac{1}{10} - \frac{1}{20} = \frac{1}{u} \Rightarrow \frac{1}{u} = \frac{- 20 - 10}{200} \Rightarrow \frac{1}{u} = \frac{- 30}{200} \Rightarrow \frac{1}{u} = \frac{- 3}{20} \Rightarrow u = - \frac{20}{3}

State whether the following statement is true or false:
A diverging mirror is used as a rear-view mirror.

Answer:

True
A diverging mirror, i.e. convex mirror, is used as a rear-view mirror, as it has a wider field of view and always forms virtual, erect and small-sized images of innumerable objects at the same time.

Page No 205:

Question 11:

What type of mirror should be used:

(a) as a shaving mirror?
(b) as a shop security mirror?

Answer:

The type of mirror that should be used

(a) as a shaving mirror is concave mirror

A ray of light of going towards the focus of a convex mirror becomes parallel to the principal axis after reflection from the mirror. Draw a labelled diagram to represent this situation.

Answer:

Answer:

Following are the ray-diagrams to show the formation of image, by a convex mirror, of the same object but in different locations:
(a) between infinity and pole of the mirror

The image will be virtual, magnified and behind the mirror when the object is between the infinity and focus of the mirror.

(b) at infinity

The image will be virtual, diminished and behind the mirror when the object is at infinity.

Page No 206:

Question 19:

The shiny outer surface of a hollow sphere of aluminium of radius 50 cm is to be used as a mirror:

(a) What will be the focal length of this mirror?
(b) Which type of spherical mirror will it provide?
(c) State whether this spherical mirror will diverge or converge light rays.

Answer:

(a) The radius of curvature of the mirror 'R' is 50 cm.

Therefore, focal length 'f' =

\frac{R}{2}

= 25 cm

(b) Since the outer surface is shiny, it will form a convex mirror.

Page No 206:

Question 20:

What is the advantage of using a convex mirror as rear-view mirror in vehicles as compared to a plane mirror? Illustrate your answer with the help of labelled diagrams.

Answer:

A convex mirror has a wider field of view, and always forms virtual, erect and small-sized images of a large number of objects at the same time. Size of the image formed by a plane mirror is equal to the size of the object; hence, it has a small field of view as compared to the convex mirror. Thus, a convex mirror is used as a rear view mirror in a car.

Following ray diagram explains it all:

2) When an object is placed extremely close to a concave mirror, a virtual, erect and large-sized image is obtained.

3) When an object is placed extremely close to a convex mirror, a virtual, erect and small-sized image is obtained.

Page No 206:

Question 25:

If a driver has one convex and one plane rear-view mirror, how would the images in each mirror appear different?

Answer:

Image formed in a plane mirror	Image formed in a convex mirror
1. Image is of the same size as the object. 2. The distance of the image from the mirror is equal to the distance of the object from the mirror. 3. It has a small field of view.	1. Image is smaller than the object. 2. Distance of the image from the mirror is not equal to the distance of the object from the mirror. 3.Iit has a large field of view.

Page No 206:

Question 26:

(a) Draw a labelled ray diagram to show the formation of image of an object by a convex mirror. Mark clearly the pole, focus and centre of curvature on the diagram.
(b) What happens to the image when the object is moved away from the mirror gradually?
(c) State three characteristics of the image formed by a convex mirror.

Answer:

(a)

(b) When the object is moved away from the mirror, the size of the image decreases.
(c) Image formed by a convex mirror is always

1) virtual
2) erect, and
3) diminished, i.e. smaller than the object

Page No 206:

Question 27:

(a) Draw a labelled ray diagram to show the formation of image in a convex-mirror when the object is at infinity. Mark clearly the pole and focus of the mirror in the diagram.
(b) State three characteristics of the image formed in this case.
(c) Draw diagram to show how a convex mirror can be used to give a large field of view.

Answer:

(a)

(b) Image formed by a convex mirror is always

1) virtual
2) erect, and

3) diminished, i.e. smaller than the object

(c) The following ray diagram shows that a convex mirror can be used to produce a large field of view. This is because when an object is in front of the convex mirror, irrespective of its distance (other than infinity), a virtual, erect and diminished image of the former is obtained. So, a convex mirror can produce image of a large number of objects at the same time.

A concave mirror is used by a dentist to examine the teeth of a person.

Page No 206:

Question 31:

If the image formed is always virtual, the mirror can be:

(a) concave or convex
(b) concave or plane
(c) convex or plane
(d) only convex

Answer:

Both convex and plane mirrors always form virtual images.

Page No 206:

Question 32:

A concave mirror cannot be used as:

(a) a magnifying mirror
(b) a torch reflector
(c) a dentist's mirror
(d) a real view mirror

Answer:

(d) a rear view mirror

A concave mirror cannot be used as a rear view mirror because it forms inverted images of distant objects.

Page No 206:

Question 33:

A boy is standing in front of and close to a special mirror. He finds the image of his head bigger than normal, the middle part of his body of the same size, and his legs smaller than normal. The special mirror is made up of three types of mirrors in the following order from top downwards:

(a) Convex, Plane, Concave
(b) Plane, Convex, Concave
(c) Concave, Plane, Convex
(d) Convex, Concave, Plane

Answer:

(c) concave, plane, convex

A concave mirror forms an image larger than the object.
Aconvex mirror forms an image smaller than the object.
A plane mirror forms an image similar to the size of the object.

Page No 207:

Question 34:

The mirror which can form a magnified image of an object is:

(a) convex mirror
(b) plane mirror
(c) concave mirror
(d) both convex and concave mirror

Answer:

A concave mirror forms a magnified image of an object.

Page No 207:

Question 35:

A real image of an object is to be obtained. The mirror required for this purpose is:

(a) convex
(b) concave
(c) plane
(d) either convex or concave

Answer:

Answer:

A man standing extremely close in front of a special mirror finds his image having a very small head, a fat body but legs of similar size. Hence,

(a) the top part of the mirror is convex, as it forms virtual, erect and small-sized image.
(b) the middle of the mirror is concave, as it forms virtual, erect and large-sized image.
(c) the bottom of the mirror is a plane mirror, as it forms virtual, erect and image of the same size as the object.

Page No 207:

Question 40:

Two big mirrors A and B are fitted side by side on a wall. A man is standing at such a distance from the wall that he can see the erect image of his face in both the mirrors. When the man starts walking towards the mirrors, he finds that the size of his face in mirror A goes on increasing but that in mirror B remains the same.

(a) mirror A is concave and mirror B is convex
(b) mirror A is plane and mirror B is concave
(c) mirror A is concave and mirror B is plane
(d) mirror A is convex and mirror B is concave

Answer:

Image formed by a plane mirror is virtual, erect and of the same size as the object.
Image formed by a concave mirror is virtual,erect and larger than the object.

Page No 209:

Question 1:

An object is kept at a distance of 5 cm in front of a convex mirror of focal length 10 cm. Calculate the position and magnification of the image and state its nature.

Answer:

Distance of the object from the mirror 'u' = -5 cm
Focal length of the mirror 'f '= 10 cm
We have to find the distance of the image 'v'.
Using the mirror formula, we get
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \Rightarrow \frac{1}{10} = \frac{1}{v} + \frac{1}{- 5} \Rightarrow \frac{1}{10} = \frac{1}{v} - \frac{1}{5} \Rightarrow \frac{1}{10} + \frac{1}{5} = \frac{1}{v} \Rightarrow \frac{1}{10} + \frac{2}{10} = \frac{1}{v} \Rightarrow \frac{1}{v} = \frac{3}{10} \Rightarrow v = \frac{10}{3} \Rightarrow v = 3.3 cm T hus, the distance of the image' v' is 3.3 cm behind the mirror . Now, using the magnification formula, we get m = \frac{- v}{u} m = \frac{- 3.3}{- 5} m = 0.66$

Thus, the image is virtual, erect and small in size.

Page No 209:

Question 2:

An object is placed at a distance of 10 cm from a convex mirror of focal length 5 cm.

(i) Draw a ray-diagram showing the formation image
(ii) State two characteristics of the image formed
(iii) Calculate the distance of the image from mirror.

Answer:

(i) Ray Diagram-

(ii) Following are the characteristics of the image formed:

(1) It is virtual and erect.
(2) It is smaller than the object.

(iii) Distance of the object 'u' = -10 cm
Focal length of the convex mirror is 5 cm.
We have to find the distance of the image 'v'.
Using the mirror formula, we get

\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \Rightarrow \frac{1}{5} = \frac{1}{v} + \frac{1}{- 10} \Rightarrow \frac{1}{5} = \frac{1}{v} - \frac{1}{10} \Rightarrow \frac{1}{5} + \frac{1}{10} = \frac{1}{v} \Rightarrow \frac{2}{10} + \frac{1}{10} = \frac{1}{v} \Rightarrow \frac{1}{v} = \frac{3}{10} \Rightarrow v = \frac{10}{3} \Rightarrow v = 3.3 cm

Thus, the distance of the image is 3.3 cm behind the mirror.

Page No 209:

Question 3:

An object is placed at a distance of 6 cm from a convex mirror of focal length 12 cm. Find the position and nature of the image.

Answer:

Distance of the object (u) = $-$ 6 cm
Focal length of the convex mirror (f) = 12 cm
We have to find the position of the image (v) and the nature of the image.
Using the mirror formula, we get:
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \Rightarrow \frac{1}{12} = \frac{1}{v} + \frac{1}{- 6} \Rightarrow \frac{1}{12} = \frac{1}{v} - \frac{1}{6} \Rightarrow \frac{1}{12} + \frac{1}{6} = \frac{1}{v} \Rightarrow \frac{1}{12} + \frac{2}{12} = \frac{1}{v} \Rightarrow \frac{1}{v} = \frac{3}{12} \Rightarrow \frac{1}{v} = \frac{1}{4} \Rightarrow v = 4 cm$

Thus, the distance of the image (v) is 4 cm behind the mirror.
Now, using the magnification formula, we get:
$m = \frac{- v}{u} m = \frac{- 4}{- 6} = \frac{2}{3} m = 0.6$

Thus, the image is virtual, erect and small.

Page No 209:

Question 4:

An object placed 20 cm in front of a mirror is found to have an image 15 cm (a) in front of it, (b) behind the mirror. Find the focal length of the mirror and the kind of mirror in each case.

Answer:

(i) Distance of the object 'u' = -20 cm

Image is formed in front of the mirror at a distance (v) of -15 cm.

Therefore, the focal length of the mirror 'f' is

\Rightarrow \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \Rightarrow \frac{1}{f} = \frac{1}{- 15} + \frac{1}{- 20} \Rightarrow \frac{1}{f} = - \frac{4}{60} - \frac{3}{60} = \frac{- 7}{60} \Rightarrow f = \frac{- 60}{7} cm

Thus, the mirror is concave and has a focal length of 60/7.

(ii) Distance of the object 'u' = -20 cm

Image is formed behind the mirror at a distance (v) of 15 cm.

Then, the focal length of the mirror 'f' will be

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \Rightarrow \frac{1}{f} = \frac{1}{15} + \frac{1}{- 20} \Rightarrow \frac{1}{f} = \frac{4}{60} - \frac{3}{60} = \frac{1}{60} \Rightarrow f = 60 cm$

Thus, the mirror is convex and has a focal length of 60 cm.

Page No 209:

Question 5:

An arrow 2.5 cm high is placed at a distance of 25 cm from a diverging mirror of focal length 20 cm. Find the nature, position and size of the image formed.

Answer:

Given,
The mirror is a diverging mirror, i.e. convex mirror.

Height of the object 'h_o'= 2.5 cm

Distance of the object from the mirror 'u' = -25 cm

Focal length of the convex mirror 'f' = 20 cm

We have to find the position of the image 'v', height of the image 'h_i' and its magnification 'm'.

Using the mirror formula, we get

\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \frac{1}{20} = \frac{1}{v} + \frac{1}{- 25} \frac{1}{20} + \frac{1}{25} = \frac{1}{v} \frac{5}{100} + \frac{4}{100} = \frac{1}{v} \frac{1}{v} = \frac{9}{100} v = \frac{100}{9} = 11.1 cm

Thus, the distance of the image 'v' is 11.1 cm behind the mirror.
Now, using the magnification formula, we get

m = \frac{- v}{u} = \frac{- 11.1}{- 25} m = 0.44

Therefore, the image is virtual, erect and smaller in size.
Again,

m = \frac{h_{i}}{h_{o}} 0.4 = \frac{h_{i}}{2.5} h_{i} = 0.44 \times 2.5 = 1.1 c m

Therefore, the height of the image is 1.1cm.

Page No 209:

Question 6:

A convex mirror used as a rear-view mirror in a car has a radius of curvature of 3 m. If a bus is located at a distance of 5 m from this mirror, find the position of image. What is the nature of the image?

Answer:

Given,
The mirror is convex.

Distance of the object from the mirror 'u' = -5 m
Radius of curvature of the mirror 'R' = 3 m

Focal length of the convex mirror 'f' =

\frac{R}{2}

= 1.5 m

We have to find the position of the image 'v', and its magnification 'm'.
Using the mirror formula, we get

\frac{1}{f} = \frac{1}{v} + \frac{1}{u}

\Rightarrow \frac{1}{1.5} = \frac{1}{v} + \frac{1}{- 5}

\Rightarrow \frac{10}{15} + \frac{1}{5} = \frac{1}{v}

\Rightarrow \frac{10}{15} + \frac{3}{15} = \frac{1}{v} \Rightarrow \frac{1}{v} = \frac{13}{15}

\Rightarrow v = \frac{15}{13} = 1.15 m

Thus, the distance of the image 'v' is 1.15 m behind the mirror.
Now, using the magnification formula, we get

m = \frac{- v}{u}

m = \frac{- (1.15)}{- 5} = \frac{115}{500} = 0.223

Thus, the image is virtual, erect and smaller in size.

Page No 209:

Question 7:

A diverging mirror of radius of curvature 40 cm forms an image which is half the height of the object. Find the object and image positions.

Answer:

Given,
The mirror is convex.

Magnification 'm' =

\frac{1}{2}

Radius of curvature 'R' = 40 cm

Focal length of the convex mirror 'f' =

\frac{R}{2}

= 20 cm

We have to find the position of the image 'v' and distance of the object from the mirror 'u'.

Using the mirror formula, we get

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} G i v e n, m = \frac{1}{2} A g a i n, m = \frac{- v}{u} T h e r e f o r e, v = - \frac{1}{2} u By putting the value of' v' in the mirror formula, we get \frac{1}{f} = \frac{1}{u} - \frac{2}{u} = \frac{- 1}{u} \Rightarrow \frac{1}{20} = \frac{- 1}{u} \Rightarrow u = - 20 c m \Rightarrow v = \frac{- (- 20)}{2} \Rightarrow v = 10 cm If an object is placed at a distance of 20 cm in front of th e mirror, the image will be formed 10 cm behind the mirror .

Page No 209:

Question 8:

The radius of curvature of a convex mirror used as a rear view mirror in a moving car is 2.0 m. A truck is coming from behind it at a distance of 3.5 m. Calculate (a) position, and (b) size of the image relative to the size of the truck. What will be the nature of the image?

Answer:

Ans

Page No 209:

Question 9:

(a) Draw a diagram to represent a convex mirror. On this diagram mark principal axis, principal focus F and the centre of curvature C if the focal length of convex mirror is 3 cm.
(b) An object 1 cm tall is placed 30 cm in front of a convex mirror of focal length 20 cm. Find the size and position of the image formed by the convex mirror.

Answer:

(a) Ray Diagram-

(b) Given,
The mirror is a diverging mirror, i.e. convex mirror.

Height of the object 'h_o'= 1 cm

Distance of the object from the mirror 'u' = -30 cm

Focal length of the convex mirror 'f' = 20 cm

We have to find the position of the image 'v', height of the image 'h_i' and its magnification 'm'.

Using the mirror formula, we get

\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \Rightarrow \frac{1}{20} = \frac{1}{v} + \frac{1}{- 30} \Rightarrow \frac{1}{30} + \frac{1}{20} = \frac{1}{v} \Rightarrow \frac{2}{60} + \frac{3}{60} = \frac{1}{v} \Rightarrow \frac{5}{60} = \frac{1}{v} \Rightarrow v = \frac{60}{5} = 12 cm

The image will be at a distance of 12 cm behind the mirror .

Now, using the magnification formula, we get

m = \frac{- v}{u} m = \frac{- 12}{- 30} = 0.4

Thus, the image is virtual, erect and smaller in size.

Also,

m = \frac{h_{i}}{h_{o}} \Rightarrow 0.4 = \frac{h_{i}}{1} \Rightarrow h_{i} = 0.4 cm

Therefore, the height of the image is 0.4 cm.

Page No 209:

Question 10:

A shop security mirror 5.0 m from certain items displayed in the shop produces on-tenth magnification.

(a) What is the type of mirror?
(b) What is the radius of curvature of the mirror?

Answer:

(a) It is a convex mirror, as it has a wide area of view.

(b) Given,
Distance of the object (u) = -5 m

Magnification (m) =

\frac{1}{10}

We have to find the position of the image (v), radius of curvature (R) and the focal length of the mirror (f).

Using the magnification formula, we get

m = \frac{- v}{u} m = \frac{- v}{- 5} \frac{1}{10} = \frac{- v}{- 5} \frac{1}{10} = \frac{v}{5} v = 0.5 m Now, using the mirror formula, we get \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \frac{1}{f} = \frac{1}{- 5} + \frac{1}{0.5} \frac{1}{f} = \frac{1}{- 5} + \frac{10}{5} \frac{1}{f} = - \frac{1}{5} + \frac{10}{5} \frac{1}{f} = \frac{9}{5} f = \frac{5}{9} R = 2 f R = \frac{10}{9}

or, R = 1.1

Therefore, the radius of curvature of the mirror (R) is 1.1 m.

Page No 210:

Question 11:

An object is placed 15 cm from (a) a converging mirror, and (b) a diverging mirror, of radius of curvature 20 cm. Calculate the image position and magnification in each case.

Answer:

Case -1

It is a converging mirror, i.e. concave mirror.

Distance of the object 'u' = -15 cm

Radius of curvature of the mirror 'R' = -20 cm

Focal length of the mirror 'f' =

\frac{R}{2}

= -10 cm

We have to find the position of the image 'v' and its magnification 'M'.

Using the mirror formula, we get

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \Rightarrow \frac{1}{- 10} = \frac{1}{- 15} + \frac{1}{v} \Rightarrow \frac{1}{v} = \frac{1}{- 10} + \frac{1}{15} \Rightarrow \frac{1}{v} = - \frac{1}{10} + \frac{1}{15} \Rightarrow \frac{1}{v} = \frac{- 15}{150} + \frac{10}{150} \Rightarrow \frac{1}{v} = \frac{- 5}{150} \Rightarrow v = - 30 cm the image will be form at a distance of 30 cm in front of converging mirror . magnification = \frac{- v}{u} m = \frac{- (- 30)}{- 15} m = - 2 magnification = - 2

Thus the image is real, inverted and large in size.

Case -2

Mirror is converging mirror i.e. convex mirror.

Distance of the object u = -15 cm

Radius of curvature of the mirror R = 20 cm

Focal length of the mirror f =

\frac{R}{2}

= 10 cm

We have to find the position of the image v = ?

Magnification = ?

U sing t h e mirror formula, w e g e t \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \Rightarrow \frac{1}{f} = \frac{1}{- 15} + \frac{1}{v} \Rightarrow \frac{1}{10} = \frac{1}{- 15} + \frac{1}{v} \Rightarrow \frac{1}{v} = \frac{1}{10} + \frac{1}{15} \Rightarrow \frac{1}{v} = \frac{15}{150} + \frac{10}{150} \Rightarrow \frac{1}{v} = \frac{25}{150} \Rightarrow \frac{1}{v} = \frac{1}{6} \Rightarrow v = 6 cm T h e r e fore, the image willl form 6 cm behind the mirror . Using the magnification formula, we get m = \frac{- v}{u} m = \frac{- 6}{- 15} m = 0.4

Thus, the image is virtual, erect and small in size.

Page No 210:

Question 12:

An object 20 cm from a spherical mirror gives rise to a virtual image 15 cm behind the mirror. Determine the magnification of the image and the type of mirror used.

Answer:

Given,
The distance of the object 'u' = -20 cm

Distance of the virtual image 'v' = 15 cm

We have to find the magnification 'm'.
Using the magnification formula, we get

m = \frac{- v}{u}

m = \frac{- 15}{- 20}

m = 0.75

Thus, the image is virtual, erect and smaller in size.
The mirror used here is convex, as it forms a smaller virtual image.

View NCERT Solutions for all chapters of Class 10