Question

An equiconvex lens of focal length $15cm$ is cut into two halves. The focal length of each half is $40cm$.

• A. True
• B. False

Answer 1

The correct option is B False

By Lens maker's formula,

$\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

After acting convex lens become plano- convex lens,

Before cutting , $R_1=R_2=R$.

$\frac{1}{15}=x[\frac{1}{R}-(\frac{-1}{R}\left)\right]$ $[x=\mu -1]$.

$\frac{x}{R}=\frac{1}{15\times 2}$-----------------1

After cut,

$\frac{1}{f_1}=x(\frac{1}{R_i}-\frac{1}{\infty })$

$\Rightarrow \frac{1}{f_1}=\frac{x}{R}$.

$\Rightarrow f_1=30cm$