Question

Image formed by the plane mirror is behind the mirror and distance of object and image from the mirror
is same if a person initially at a distance of 20 cm from the mirror moves away from one mirror at the
rate of 5 cm/s. Find the distance between image and object at the end of 10 seconds:
(A) 20 cm (B) 100 cm (C) 120 cm (D) 140 cm​

Answer 1

Hi !

The initial distance between image and object = 40cm (20 x 2)

5 cm/s , 10 sec so = 10 x 5= 50 cm (50 x 2=100 cm )

100 +40 = 140 cm

D is the answer

Answer 2

Given that,

Distance = 20 cm

Time = 10 sec

The mirror moves away from one mirror at the  rate of 5 cm/s.

The distance of object and image from the mirror  is same.

We need to calculate the initial distance between object and image

Using formula of distance

$D=2\times \text{distance of object}$

Put the value into the formula

$D_{i}=2\times20$

$D_{i}=40\ cm$

We need to calculate the distance of the object after move

Using formula of distance

$D= v \times t$

$D=5\times10$

$D=50\ cm$

The distance between object and image after moved

$D_{f}=2D$

$D_{f}=2\times50$

$D_{f}=100\ cm$

We need to calculate the total distance

$D'=D_{i}+D_{f}$

Put the value into the formula

$D'=40+100$

$D'=140\ cm$

Hence, The distance between object and image is 140 cm.

(D) is correct option.