Question

A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter 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:

• A. \(f\) and \(\dfrac{I}{4}\)
• B. \(\dfrac{3f}{4}\) and \(\dfrac{I}{2}\)
• C. \(f\) and \(\dfrac{3I}{4}\)
• D. (f)/(2) and (I)/(2)

Hint:

Blocking part of a lens reduces brightness, not image size or focal length.

Answer 1

The Correct Option is C

Step 1: Focal length depends only on curvature and refractive index.
Step 2: Effective transmitting area reduces by 1/4, so remaining area is 3/4.
Step 3: Hence intensity: I'=(3I)/(4)