Ray Optics and Optical Instruments - Result Question 35
35. 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. Focal length of lens and intensity of image now will be respectively:
[2010]
(a) $f$ and $\frac{I}{4}$
(b) $\frac{3 f}{4}$ and $\frac{I}{2}$
(c) $f$ and $\frac{3 I}{4}$
(d) $\frac{f}{2}$ and $\frac{I}{2}$
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Answer:
Correct Answer: 35. (c)
Solution:
- (c) By covering aperture, focal length does not change. But intensity is reduced by $\frac{1}{4}$ times, as aperture diameter $\frac{d}{2}$ is covered.
$\therefore \quad I^{\prime}=I-\frac{I}{4}=\frac{3 I}{4}$
$\therefore \quad$ New focal length $=f$ and intensity $=\frac{3 I}{4}$.
