SATHEE: Optics Question 606

Optics Question 606

Question: 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 or light is

[CBSE PMT 1995]

Options:

A) $\frac{A_{1} + A_{2}}{2}$

B) ${[\frac{1}{A_{1}} + \frac{1}{A_{2}}]}^{- 1}$

C) $\sqrt{A_{1} A_{2}}$

D) ${[\frac{\sqrt{A_{1}} + \sqrt{A_{2}}}{2}]}^{2}$

Show Answer

Answer:

Correct Answer: C

Solution:

$m_{1} = \frac{A_{1}}{O}$ and $m_{2} = \frac{A_{2}}{O}$

$\Rightarrow m_{1} m_{2} = \frac{A_{1} A_{2}}{O_{2}}$ .

Also it can be proved that $m_{1} m_{2} = 1$ .

So $O = \sqrt{A_{1} A_{2}}$

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Optics Question 606

Question: A lens is placed between a source of light and a wall. It forms images of area A1 and A2 on the wall for its two different positions. The area of the source or light is

Options:

Answer:

Solution:

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Question: 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 or light is