Question

A thin lens has focal length $f$, and its aperture has diameter $d$. It forms an image of intensity $I$. If the central part of the aperture, of diameter $\frac{d}{2}$, is blocked by an opaque paper, the focal length of the lens and the intensity of image will become

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

Answer 1

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

Blocking a lens by black paper does not change the geometric configuration of the lens, hence the rays going to the uncovered region refract as they did before the lens being covered. Hence the focus of the lens does not change.

However the intensity would be reduced.

Area of lens of diameter $D=\pi D^2/4$

Are of lens covered $=\pi (\frac{D}{2}{)}^2/4=\pi D^2/16$

Fraction of area uncovered $=0.75$

Since intensity is proportional to the area of lens, the intensity of image obtained after covering$=\frac{3I}{4}$