SATHEE: Optics Question 166

Optics Question 166

Question: A circular disc is placed in front of a narrow source. When the point of observation is 2 m from the disc, then it covers first HPZ. The intensity at this point is I. When the point of observation is 25 cm from the disc then intensity will be

Options:

A) ${(\frac{R_{6}}{R_{2}})}^{2} I$

B) ${(\frac{R_{7}}{R_{2}})}^{2} I$

C) ${(\frac{R_{8}}{R_{2}})}^{2} I$

D) ${(\frac{R_{9}}{R_{2}})}^{2} I$

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Answer:

Correct Answer: D

Solution:

$I = \frac{R_{2}^{2}}{4}$ given $n_{1} b_{1} = n_{2} b_{2}$

Therefore $1 \times 200 = n_{2} \times 25$

$∴ n_{2} = 8 H P Z$

$∴ I = {(\frac{R_{9}}{2})}^{2}$

$= {(\frac{R_{9}}{R_{8}} \times \frac{R_{8}}{R_{7}} \times \frac{R_{7}}{R_{6}} \times \frac{R_{6}}{R_{5}} \times \frac{R_{5}}{R_{4}} \times \frac{R_{4}}{R_{3}} \times \frac{R_{3}}{R_{2}} \times \frac{R_{2}}{R_{2}})}^{2}$

$= {(\frac{R_{9}}{R_{2}})}^{2} I$

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Optics Question 166

Question: A circular disc is placed in front of a narrow source. When the point of observation is 2 m from the disc, then it covers first HPZ. The intensity at this point is I. When the point of observation is 25 cm from the disc then intensity will be

Options:

Answer:

Solution:

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