Question

A lens having focal length $f$ and aperture of diameter $d$ forms an image of intensity $I$. Aperture of diameter $\frac{d}{2}$ in central region of lens is covered by a black paper. Now Focal length of lens and intensity of image will be respectively

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

Answer 1

The correct option is C $f\text{and}\frac{3I}{4}$

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

Thus, if the material of the lens and surrounding media is not changed, the focal length of the lens will only depend on curvature of the surface.

$\because R_1,R_2$ are not changed.

So,
$f_{\text{initial}}=f_{\text{final}}=f$

Also, $I\propto A$

$A=$ Area exposed to incident light
$I=$ intensity of image.

$\Rightarrow \frac{I_2}{I_1}=\frac{A_2}{A_1}$

Initial area, $A_1=\pi \frac{d^2}{4}$

Since half of the aperture is covered by a black paper.

$A^{\prime }=\pi {\left(\frac{d/2}{2}\right)}^2$

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

After blocking, the exposed area becomes
$\Rightarrow A_2=\pi \frac{d^2}{4}-\pi \frac{d^2}{16}$

$\Rightarrow A_2=\pi \frac{3d^2}{16}$

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

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

Hence, option (c) is the correct answer.

Why this question? Bottom line : If aperture of lens is obstructed then focal length of lens remains same but intensity of image decreases.