Question

A double convex lens of focal length f is cut into 4 equivalent parts. One cut is perpendicular to the axis and the other is parallel to the axis of the lens. The focal length of each part is :

• A. $f/2$
• B. $f$
• C. $2f$
• D. $4f$

Answer 1

The correct option is C $2f$

By lens maker formula.

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

So, cutting along the axis does not change the focal length but cutting perpendicular to the axis change radius of curvature of one side so, focal length changes.

So, for a double convex lens, we get $R_1=R_2=R$

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

$\Rightarrow \frac{1}{f}=(\mu -1)\left[\frac{2}{R}\right]\Rightarrow f=\frac{R}{2(\mu -1)}$

Now as it is cut perpendicular to axis $R_2\to 0$

So, $\Rightarrow \frac{1}{f}=(\mu -1)[\frac{1}{R}-0]$

$\Rightarrow \frac{1}{f_1}=(\mu -1)\frac{1}{R}\Rightarrow f_1=\frac{R}{(\mu -1)}$

$\therefore f_1=2f$

So, focal length of each part as $2f$.