Question

A thin equiconvex lens of focal length f(in air) and refractive index m has thickness t. How much thick it will appear to an observer as shown in figure?

• A. $\frac{t}{\mu }$
• B. $\frac{2ft}{2\mu f-t}$
• C. t
• D. None

Answer 1

The correct option is B

$\frac{2ft}{2\mu f-t}$

Let R is the radius of curvature of two surfaces of lens

$\therefore \frac{1}{f}=(\mu -1)\frac{2}{R}$

For the observer the points A seems to be at its actual position but B at over position for Refraction :

$\mu =-t,{\mu }_1=\mu ,{\mu }_2=1,R=-R$

$\therefore \frac{1}{v}-\frac{\mu }{-t}=\frac{1-\mu }{-R}\Rightarrow \frac{1}{v}=\frac{\mu -1}{R}-\frac{\mu }{t}=\frac{1}{2f}-\frac{\mu }{t}\therefore v=\frac{2ft}{t-2\mu f}=-\frac{2ft}{2\mu f-t}$