Question

A thin lens focal length $f _ { 1 }$ and its aperture has diameter d. It forms an image of intensity I. Now the central part of the aperture upto diameter d/2 is blocked by an opaque paper. The focal length and image intensity will change to

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

Answer 1

Answer: (D)

$f$ and $\frac { 3 I } { 4 }$

Answer 2

Centre part of the aperture up to diameter $\left( A = \frac { \pi d ^ { 2 } } { 4 } \right)$. Hence remaining area . Also, we know that intensity ∝ Area

⇒ $\frac { I ^ { \prime } } { I } = \frac { A ^ { \prime } } { A } = \frac { 3 } { 4 }$ ⇒ $I ^ { \prime } = \frac { 3 } { 4 } I$.

Focal length doesn't depend upon aperture.