A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter d2 in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively
- f and I4
- 3f4 and I2
- f and 3I4
- f2 and I2
Answer
(3) f and 3I4
Solution
Focal length depends only on the radii of curvature and the refractive index it stays f.
Intensity of the image light-transmitting area of the lens.
Area blocked = (d/2)24 = 14 d24
Area still open = 34 of the original
New intensity = 3I4
Answer: (3)