Question

A thin lens has a focal length $f$ and its aperture has a diameter $d$ . It 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 would change to

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

Answer 1

Answer: (D)

The focal length remains $f$ , as any small portion of the lens forms a complete image of the object and has the same focal length.
While the intensity of the image depends on the area as,
$Intensity \propto \left(A r e a\right)^{2}$
$\frac{I_{2}}{I_{1}}=\frac{A_{2}}{A_{1}}$
$\frac{I_{2}}{I}=\frac{\frac{\pi d^{2}}{4} - \frac{\pi \left(\frac{d}{2}\right)^{2}}{4}}{\frac{\pi d^{2}}{4}}$
$\frac{I_{2}}{I}=\frac{3}{4}$
$I_{2}=\frac{3}{4}I$