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.