Question

A beam of light is incident at point 1 on a screen, a plane glass plate of thickness t and refractive index n is placed in the path of light. The displacement of point will be :

• A. $t(1-\frac{1}{n})nearer$
• B. $t(1+\frac{1}{n})nearer$
• C. $t(1-\frac{1}{n})farther$
• D. $t(1+\frac{1}{n})father$

Answer 1

The correct option is A $t(1-\frac{1}{n})nearer$

Consider refraction of the point from surface 1,

$\frac{n}{v_1}-\frac{1}{-x}=\frac{n-1}{\infty }$

Hence $v_1=-nx$

This image acts as object for the refraction through surface 2. For this refraction,

$u_2=-nx-t$

Thus $\frac{1}{v_2}-\frac{n}{-nx-t}=\frac{1-n}{\infty }$

$v_2=-x-\frac{t}{n}$

Thus the image is at a distance of $x+\frac{t}{n}$ to the left of surface 2.

Therefore the image is at a distance of $x+\frac{t}{n}-t$ to the left of surface 1.

Original object was at a distance of $x$ to the left of surface 1.

Thus the final image is to the right of object by a distance of $x-(x+\frac{t}{n}-t)=t(1-\frac{1}{n})$ nearer.