question_answer

A thin lens has focal length f, and its aperature has diameter d It forms an image of intensity \[I\] Now the central part of the aperature upto  diameter d/2 is blocked by an opaque paper. The focal length and image intensity will change to

A)  \[\frac{f}{2}and\frac{I}{2}\]       

B)         \[f\,and\frac{I}{4}\]

C)  \[\frac{3f}{4}\,and\frac{I}{2}\] 

D)         \[f\,and\frac{3I}{4}\]  

Correct Answer: D

Solution :

 If the area of lens of diameter \[d\] is\[A\], then the area of the central opaque part is\[\frac{A}{4}\], so area of the remaining transparent part is\[\frac{3A}{4}\]. So, the amount of light entering the glass is three-fourth of initial. Amount of light energy means, intensity. Focal length of a lens depends only on\[\mu ,\,\,{{R}_{1}},\,\,{{R}_{2}}\]. Since these are constant, focal length will remain same.