Question

A thin lens has focal length $f$, and its aperture has diameter $d$. If the central part of the aperture, of diameter $d/2$ is blocked by an opaque paper, the focal length of the lens and the intensity of image will become :

• A. $\frac{f}{2},\frac{I}{2}$
• B. $f,\frac{I}{4}$
• C. $\frac{3f}{4},\frac{I}{2}$
• D. $f,\frac{3I}{4}$

Answer 1

The correct option is D $f,\frac{3I}{4}$

Focal length $=F$

Aperture diameter $=d$

Diameter of portion of lens to be blocked $=\frac{d}{2}$

As the focal length is only dependent on the sphere from which lens is made so it does not change

Initial area of uncovered lens,$A_1=\pi {\left(\frac{d}{2}\right)}^2$

$\Rightarrow A_1=\pi \frac{d^2}{4}$

Area of lens covered by opaque paper,$A_2=\pi {\left(\frac{d}{4}\right)}^2$

$\Rightarrow A_2=\pi \frac{d^2}{16}$

Remaining uncovered area,$A_3=A_1-A_2$

$\Rightarrow A_3=\frac{3\pi d^2}{16}$

Now as, $I\propto A_{uncovered}$

$\Rightarrow \frac{I_2}{I_1}=\frac{A_3}{A_1}=\frac{\frac{3\pi d^2}{16}}{\frac{\pi d^2}{4}}$

$\Rightarrow \frac{I_2}{I_1}=\frac{3}{4}$

$I_2=\frac{3}{4}{I}_1$

Hence should be $f,\frac{3}{4}{I}_1$