Question

An object is placed 20cm in front of a convex mirror and the image is formed 10 cm from the

mirror. Calculate new image distance if object is moved 5cm towards the mirror​

Answer 1

Distance of object from mirror = 20cm

Distance of image from mirror = 10cm

Let focal length of mirror be f.

According to the question :-

$\rm \longrightarrow\dfrac{1}{10} + \dfrac{1}{( - 20)} = \dfrac{1}{f} \\ \\ \\ \rm \longrightarrow\dfrac{1}{10} - \dfrac{1}{20} = \dfrac{1}{f} \\ \\ \\ \rm \longrightarrow \dfrac{2 - 1}{20} = \dfrac{1}{f} \\ \\ \\ \rm \longrightarrow \dfrac{ 1}{20} = \dfrac{1}{f} \\ \\ \\ \\ \\ \rm \longrightarrow \: 20 \: cm = f$

Therefore, focal length of mirror = 20 cm

Now, if object is moved 5cm towards the mirror then distance of object from mirror = 20cm - 5 cm

= 15 cm

Let new image distance be v.

According to the question :-

$\rm \longrightarrow \dfrac{1}{v} + \dfrac{1}{( - 15)} = \dfrac{1}{20} \\ \\ \\ \rm \longrightarrow \dfrac{1}{v} = \dfrac{1}{20} + \dfrac{1}{ 15}\\ \\ \\ \rm \longrightarrow \dfrac{1}{v} =\dfrac{3 + 4}{60}\\ \\ \\ \rm \longrightarrow \dfrac{1}{v} =\dfrac{7}{60}\\ \\ \\ \rm \longrightarrow {v} =\dfrac{60}{7}cm$

Therefore , new image distance = 60/7 = 8.57 cm

Answer 2

(i) For uniform motion of an object, its distance-time graph is a straight line with constant slope. (ii) For non-uniform motion of an object, its distance-time graph is a curved line with increasing or decreasing slope.