Solved Examples of Light Reflection

Solved Examples of Light Reflection

1. An object is placed in front of a plane mirror. If the mirror is moved away from the object through a distance x, by how much distance will the image move?

Solution : Suppose the object O was initially at a distance d from the plane mirror M as shown in fig. The image formed at O’ is at a distance d behind the mirror. Now, the mirror is shifted by a distance x to M’ such that the distance of the object from M’ becomes d + x. The image now formed at O” which is also at a distance d + x from M’.

Light Reflection SE1
Light Reflection SE1

So, OM = MO’ = d

OM’ = M’O” = d + x

Thus,    OO” = OM’ + M’O” = 2(d + x) …(1)

when     OO’ = OM +  MO’ = 2d             …(2)

∴        O’O” = OO” – OO’

= 2(d + x) – 2d

= 2x

Thus, the image is shifted from O’ to O” by a distance 2x.

 

2. An insect is at a distance of 1.5 m from a plane mirror. Calculate the following?

(i) Distance at which the image of the insect is formed.

(ii) distance between the insect and its image.

Solution : (i) The distance of insect from the mirror = 1.5 m

∴ The distance of insect from the mirror is also equal to 1.5 m. The image is formed at 1.5 m behind the mirror.

(ii) The distance between the insect and   image = 1.5 + 1.5 = 3 m

3. A concave mirror is made up by cutting a portion of a hollow glass sphere of radius 30 cm. Calculate the focal length of the mirror.

Solution : The radius of curvature of the mirror = 30 cm

Thus, the focal length of the mirror

= \frac{{30 cm}}{2}$= 15 cm

4. An object is placed at a distance of 15 cm from a concave mirror of focal length 10 cm. Find the position of the image.

Light Reflection SE4
Light Reflection SE4

Solution : We have u = –15 cm and f = –10 cm

Using the relation,,  \frac{1}{v} + \frac{1}{u} = \frac{1}{f}$  we get

\frac{1}{v} + \frac{1}{{-15}} =\frac{1}{{-10}}$

or                =   – = –

or               v = –30 cm

So the image will be formed 30 cm from the mirror. Since n has a negative sign, the image is formed to the left of the mirror, i.e. in front of the mirror as shown in fig.

 

Ex.5     A 3 cm long object is placed perpendicular to the principal axis of a concave mirror. The distance of the object from the mirror is 15 cm, and its image is formed 30 cm from the mirror on the same side of the mirror as the object . Calculate the height of the image formed.

Sol.       Here u = –15 cm and n = –30 cm

Size of the object, h = 2 cm

Magnification,    m = m =

or                     = – = 2

or                 h’ = – 2 × h = – 2 × 3

= – 6 cm

So the height of the image is 6 cm. The minus sign shows that it is on the lower side of the principal axis, i.e. the image is inverted.

 

Ex.6     A 1.4 cm long object is placed perpendicular to the principal axis of a convex mirror of focal length 15 cm at a distance of 10 cm from it. Calculate the following :

(i) location of the image

(ii) height of the image

(iii) nature of the image

 

Sol. (i)   For a convex mirror, focal length is positive.

Therefore, f = +15 cm and u = –10 cm

Using the relation, +=, we get

+=

or                      = += =

or                     n = 6 cm

Since n is positive, the image is formed to the right of the mirror at a distance 6 cm from it.

 

(ii) Magnification,

or               m = = –

= = + 0.6

or               h’ = + 0.6 × h0  = 0.6 × 1.4

= 0.84 cm.

Thus, the height of the image is 0.84 cm.

(iii) Since h’ is positive, the image will be on the same side of the principal axis as the object.

Hence, the image is virtual, erect and diminished.

 

Ex.7     An object is placed at a distance of 40 cm from a convex mirror of focal length 30 cm. Find the position of image and its nature.

 

Sol.       Here, object distance, u = –40 cm

Focal length of convex mirror, f = +30 cm

Now, using mirror formula,  +  =  we get

+ =

or                =  +  =

or               n =

The positive sign shows that the image is formed on the right, i.e. behind the mirror.

Now, magnification,

Þ m = –  = –  = +

Since, the magnification is positive, the image is erect. Thus, the image is formed 17.1 cm behind the mirror. The image is virtual, erect and diminished.

Ex.8     A 3 cm high object is placed at a distance of
30 cm from a concave mirror. A real image is formed 60 cm from the mirror. Calculate the focal length of the mirror and the size of the image.

Sol.       Object distance,        u = –30

Image distance,        n = –60

(real image is formed on the same side)

Now, using the mirror formula, +=     we get

+  =

or  = – = –

or   f = – 20 cm

\ Focal length of the mirror = 20 cm

Magnification    m =  = –

or                      = –

or                     h’ = 3 × (–2) = –6 cm

The height of the image is 6 cm. The negative sign shows that the image is inverted.

 

Ex.9     A 1 cm high object is placed at 20 cm in front of a concave mirror of focal length 15 cm. Find the position and nature of the image.

Sol.       u = –20 cm, f = –15 cm, h0 = 1 cm

Using mirror formula, += we get

+  =

or                     = – + = –

\                     n = – 60 cm

The image is formed 60 cm from the mirror. Since, the signs of u and n are the same, the object and image are formed on the same side of the mirror. Therefore, the image is real.

Now magnification,

m ==  = = –3

\    h’ = –3h = – 3 × 1 cm = – 3cm

The negative sign shows that the image is inverted. Thus, the image is real, inverted and of size 3 cm and formed 60 cm in front of the mirror.

 

Ex.10    An object 4 cm high is placed 25 cm in front of a concave mirror of focal length 15 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and size of image.

Sol.       Here, u = –25 cm, f = –15 cm, h = + 4 cm

Using the mirror formula,+=, we get

or         + =

or         = – = –+= –

or         n =  = –37.5 cm

Thus, the screen must be placed 37.5 cm from the mirror on the same side as the object.

Now, magnification, m = =

= – 1.5

or   h’ = – 1.5 × 4 = –6 cm

Negative sign shows that the image is inverted. Hence, the image is real, inverted and of size 6 cm.

 

Ex.11    An object 5 cm high is placed at a distance of
20 cm from a convex mirror of radius of curvature 30 cm. Find the position, nature and size of image.

Sol.       Here, u = –20 cm, h = 5 cm

Radius of curvature, r = +30 cm

\ Focal length,  f =  = + = + 15 cm

Using the mirror formula, +=, we get

+ =

or               =+ =

or                     n = cm

The image is formed 8.5 cm from the mirror. The positive sign shows that the image is formed on the other side or behind the mirror. So the image is virtual.

Magnification,   m = = –

or   = –  = + =

or                     h’ = 5 × =cm

 

The height of the image is 2.1 cm. Positive sign shows that the image is erect.

 

Ex.12    A convex mirror used on a automobile has 3 m radius of curvature. If a bus is located 5 m from this mirror, find the position, nature and size of image.

 

Sol.       Here, u = –5 m, r = +3m

\               f = = +  1.5 m

Using the relation,  + =, we get

or           + =

or         = +  = + 1.15 m

The image is 1.15 m behind the mirror.

Magnification, m == –= – = + 0.23

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