KVPY Sample Paper KVPY Stream-SX Model Paper-29

A lens is placed between a source of light and a wall. It forms images of area \[{{A}_{1}}\] and \[{{A}_{2}}\] on the wall, for its two different positions. The area of the source of light is (source and wall are fixed)-

A) \[{{({{A}_{1}}{{A}_{2}})}^{1/2}}\]

B) \[\frac{{{A}_{1}}+{{A}_{2}}}{2}\]

C) \[{{\left( \frac{1}{{{A}_{1}}}+\frac{1}{{{A}_{2}}} \right)}^{-1}}\]

D) \[{{\left( \frac{\sqrt{{{A}_{1}}}+\sqrt{{{A}_{2}}}}{2} \right)}^{2}}\]

Correct Answer: A

Solution :

From Nerwton's equation of lens

size of object = \[{{O}^{2}}={{l}_{1}}{{l}_{2}}\]

where \[{{l}_{1}}\] is size of image of object and \[{{l}_{2}}\]is size of image when positions of object & image are interchanged

So \[{{A}^{2}}={{A}_{1}}{{A}_{2}}\Rightarrow \mathbf{A}=\sqrt{{{A}_{1}}{{A}_{2}}}\]