Question

The given lens is broken into four parts and rearranged as shown. If the initial focal length is f and initial intensity of image is I, then after rearrangement the equivalent focal length and intensity of image is?

• A. f/2, I/4
• B. f/2, I/2
• C. f/4, I/4
• D. f/4, I/2

Answer 1

Answer: (B)

f/2, I/2

Answer 2

Each part has a focal length = 2f

\ = $\frac { 1 } { \mathrm { f } _ { 1 } }$ +++=$\frac { 1 } { 2 \mathrm { f } }$+$\frac { 1 } { 2 \mathrm { f } }$+$\frac { 1 } { 2 \mathrm { f } }$+$\frac { 1 } { 2 \mathrm { f } }$=

\ F =

Since the area of the lens exposed to incident light is half of the original and since intensity of image is proportional to area of the lens.

\ intensity of image becomes half of the initial

\ I¢ = I/2

Therefore the answer will be (2).