Question

Width of a slab is 6cm whose meu=3/2. If its rear surface is silvered and object is placed at a distance 28cm from the front face. Calculate the final position of image from the silvered surface???

Answer 1

The final position of image is 30 cm from the silvered surface.

Solution:

The slab produces an image at a distance of let us say x with shifting of the object placed in front of it with the respect to the direction of the incidence of light. Thereby the final position of the image is given as follows,

As we know that the image produced by slab is given by, $\bold{x=t\left[1-\frac{1}{\mu}\right]}$

$\begin{array}{l}{x=6\left[1-\frac{1}{\frac{3}{2}}\right]} \\ {x=6\left[1-\frac{2}{3}\right]} \\ {x=6\left[\frac{1}{3}\right]} \\ {x=\frac{6}{3}} \\ {x=2 \mathrm{cm}}\end{array}$

The distance of the object from the virtual plane mirror will be 34 – x = 34 – 2 = 32 cm

Virtual plane mirror makes image at a distance of 32 cm behind the actual mirror.  

Distance of image from actual mirror = Image distance – Shift image distance = 32 – 2 = 30 cm

Answer 2

Explanation:

see this image , and remember one thing that THE DISTANCE OF OBJECT FROM PLANE MIRROR IS SAME AS DISTANCE OF IMAGE FROM PLANE MIRROR (in this question its a virtual plane mirror , but the concept is same )