Physics - Geometrical Optics Question with Solution | TestHub

For the initial equiconvex lens, the radii of curvature are $R_1=R$ and $R_2=-R$.

Using the lens maker's formula, the initial focal length $f$ is given by

$\frac{1}{f}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=(\mu -1)\left(\frac{1}{R}-\frac{1}{-R}\right)=\frac{2(\mu -1)}{R}$

The initial power is $P=\frac{1}{f}=\frac{2(\mu -1)}{R}$.

When one surface is made plane by grinding, its radius of curvature becomes infinity. Let the new radii be $R_1^{\prime }=R$ and $R_2^{\prime }=\infty$.

The new focal length $f^{\prime }$ is

$\frac{1}{f^{\prime }}=(\mu -1)\left(\frac{1}{R}-\frac{1}{\infty }\right)=\frac{\mu -1}{R}$

Comparing the two equations, we get

$\frac{1}{f^{\prime }}=\frac{1}{2}\left(\frac{2(\mu -1)}{R}\right)=\frac{1}{2f}$

$\Rightarrow f^{\prime }=2f$

The new power $P^{\prime }$ is

$P^{\prime }=\frac{1}{f^{\prime }}=\frac{1}{2f}=\frac{P}{2}$

Answer: $2f$ and $\frac{P}{2}$