Question

A thin lens focal length f and its aperture has diameter d. It forms an image of intensity I. Now the central part of the aperture upto diameter is blocked by an opaque paper. The focal length and image intensity will change to –

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

Answer 1

Answer: (D)

f and

Answer 2

I ∝ A² Ž $\frac { \mathrm { I } _ { 2 } } { \mathrm { I } _ { 1 } }$ = = =$\frac { 3 } { 4 }$

$\frac { \mathrm { I } _ { 2 } } { \mathrm { I } _ { 1 } }$ = $\frac { 3 } { 4 }$ Ž I₂ = $\frac { 3 } { 4 }$I₁

And focal length remains unchanged.