Suppose a ball is placed in front of a concave mirror and a real image that is twice the size of the ball is formed on a screen. The ball and the screen are then moved until the image is five times the size of the object. If the shift of the screen is d, then the shift in the object is:​

Correct option is B

​The correct answer is (b) $$\frac{d}{10}$$

​Explanation:

A real image that is twice the size of the ball (object) is formed on the screen.

Magnification (M) = -2, because the image is twice the size of the object and is real (real images are inverted).

The magnification is related to the object and image distances by the formula: M=- $$\frac{v}{u}$$​​

where: is the image, distance is the object distance, M is the magnification 

For the initial situation, where, M= -2

-2= $$\frac{v1}{u1}$$,

thus, v₁ = 2u₁

So, the image distance is twice the object distance.

After shifting the object and screen, the magnification becomes -5 (image is five times the size of the object).

For the final situation:

-5=-$$\frac{v2}{u2}$$​​

v₂= 5 u₂

So, the image distance is five times the object distance.

​Let the shift of the screen be denoted as 'd'. Therefore, we have:

d=2u₁

So, u₁= $$\frac{d}{2}$$

Therefore,  change in object distance (u)= $$\frac{d}{10}$$​​

The shift in the object is d/10.​