Question
A lens having focal length $$f$$ and aperture of diameter $$d$$ forms an image of intensity $$I.$$ Aperture of diameter $$\frac{d}{2}$$ in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively
A.
$$f\,{\text{and}}\,\frac{I}{4}$$
B.
$$\frac{{3f}}{4}\,{\text{and}}\,\frac{I}{2}$$
C.
$$f\,{\text{and}}\,\frac{{3I}}{4}$$
D.
$$\frac{f}{2}\,{\text{and}}\,\frac{I}{2}$$
Answer :
$$f\,{\text{and}}\,\frac{{3I}}{4}$$
Solution :
As we know that
Intensity, $$I \propto A$$ (Area exposed)
$$\eqalign{
& \Rightarrow \frac{{{I_2}}}{{{I_1}}} = \left[ {\frac{{{A_2}}}{{{A_1}}}} \right] = \frac{{\frac{{\pi {d^2}}}{4} - \frac{{\frac{{\pi {d^2}}}{4}}}{4}}}{{\frac{{\pi {d^2}}}{4}}} = \frac{3}{4} \cr
& \Rightarrow {I_2} = \frac{3}{4}{I_1} \cr} $$
and focal length remains unchanged.