Question

A thin lens focal length $f_1$ and its aperture has diameter d. It forms an image of intensity I. Now the central part of the aperture upto diameter $\frac{d}{2}$ is blocked by an opaque paper. The focal length and image intensity will change to

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

Answer 1

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

Centre part of the aperture up to diameter $\frac{d}{2}$ is blocked i.e. $\frac{1}{4}$th area is blocked $(A=\frac{\pi d^2}{4})$. Hence remaining area $A^{\prime }=\frac{3}{4}A$. Also, we know that intensity $\propto$ Area $\Rightarrow \frac{I^{\prime }}{I}=\frac{A^{\prime }}{A}=\frac{3}{4}\Rightarrow I^{\prime }=\frac{3}{4}I$.
Focal length doesn't depend upon aperture.