Question

A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter in central region of lens is covered by a black paper. Focal length of lens intensity of image now will be respectively;

• A.
• B.
• C. f and
• D.

Answer 1

Answer: (C)

f and

Answer 2

: Facal length of the lens remains same. Intensity of image formed by lens is proportional to area exposed to incident light from object.

or

Initial area, $\mathrm { A } _ { 1 } = \pi \left( \frac { \mathrm { d } } { 2 } \right) ^ { 2 } = \frac { \pi \mathrm { d } ^ { 2 } } { 4 }$

After blocking exposed area,

$\mathrm { A } _ { 2 } = \frac { \pi ( \mathrm { d } / 2 ) ^ { 2 } } { 4 } = \frac { \pi \mathrm { d } ^ { 2 } } { 4 } - \frac { \pi \mathrm { d } ^ { 2 } } { 16 } = \frac { 3 \pi \mathrm {~d} ^ { 2 } } { 16 }$

$\therefore \frac { \mathrm { I } _ { 2 } } { \mathrm { I } _ { 1 } } = \frac { \mathrm { A } _ { 2 } } { \mathrm {~A} _ { 1 } } = \frac { \frac { 3 \pi \mathrm {~d} ^ { 2 } } { 16 } } { \frac { \pi \mathrm {~d} ^ { 2 } } { 4 } } = \frac { 3 } { 4 }$

or $\mathrm { I } _ { 2 } = \frac { 3 } { 4 } \mathrm { I } _ { 1 } = \frac { 3 } { 4 } \mathrm { I } \quad \left( \because \mathrm { I } _ { 1 } = \mathrm { I } \right)$

Hence, focal length of lens = f, intensity of the image