question_answer

The focal length of a thin lens is \[f\] and the diameter of its aperture is d. This makes an image of brightness \[I\]. If the central portion of the aperture be covered upto d/2 diameter by an opaque paper, then the focal length and the brightness of image would respectively be

A)  \[f\text{/2}\,\text{and}\,\text{I/2}\]                    

B)  \[f\,\text{and}\,\text{I/4}\]

C)  \[f\text{/2}\,\text{and}\,\text{I/4}\]                                    

D)  \[f\,\text{and}\,3\text{I/4}\]

Correct Answer: D

Solution :

Covering the aperture of a lens f will not change but luminous intensity \[I=k(\pi {{r}^{2}})\] Uncovered area \[=\pi {{r}^{2}}-\pi {{\left( \frac{r}{2} \right)}^{2}}=\frac{3}{4}\pi {{r}^{2}}\] So, new intensity \[=\frac{3}{4}k\pi {{r}^{2}}\]                                 \[=\frac{3}{4}I\]