Question

A thin lens has a focal length $f_{1}$ and its aperture has a diameter $d$ . If forms an image of intensity $I$ . Now, the central part of the aperture up to 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. $\frac{3 f}{4}$ and $\frac{I}{2}$
• D. $f$ and $\frac{3 I}{4}$

Answer 1

Answer: (D)

The focal length of the lens remains unchanged by changing its aperture.
The intensity of the image is proportional to the uncovered area.
$\frac{I_{\text{f}}}{I_{\text{i}}}=\frac{A_{\text{f}}}{A_{\text{i}}}=\frac{\frac{\pi }{4} \left[d^{2} - \frac{d^{2}}{4}\right]}{\frac{\pi }{4} d^{2}}=\frac{3}{4}$
$I_{\text{f}}=\frac{3}{4}I_{\text{i}}=\frac{3}{4}I$