Question

A thin lens of focal length f has a diameter d of its aperture and forms an image of intensity I. Now the central part of the aperture up to a diameter $\frac{d}{2}$ is blocked by white paper. Then

• A. the focal length changes to $\frac{f}{2}$
• B. the focal length remains the same as $f$
• C. the intensity of the image changes to $\frac{3I}{4}$
• D. the intensity of the image changes to $\frac{I}{4}$

Answer 1

Answer: (B), (C)

The correct options are
B the focal length remains the same as $f$
C the intensity of the image changes to $\frac{3I}{4}$

$Intensityisproportionaltoarea.BlockingtheD/2,thearealeftis3/4A.$

$I=I_0\times d^2$

$I=I_0\times (d^2-(d/2{)}^2)$

$I=I_0\times \frac{3d^2}{4}=\frac{3I}{4}$

Focal length of a lens doesn't depend on the aperture