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Q.

A circular disc is placed in front of a narrow source. When the point of observation is at a distance of 1 meter from the disc, then the disc covers first HPZ. The intensity at this point is I₀. The intensity at a point distance 25 cm from the disc will be

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a

$I_{1} = 5 .03 I_{0}$

b

$I_{1} = 53 I_{0}$

c

$I_{1} = 0 .053 I_{0}$

d

$I_{1} = 0 .531 I_{0}$

answer is A.

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Detailed Solution

$I_{0} = R^{2} = \frac{R_{2}^{2}}{4} Number of HPZ covered by the disc at b = 25 cm n_{1} b_{1} = n_{2} b_{2} n_{2} = \frac{n_{1} b_{1}}{b_{2}} = \frac{1 \times 1}{0.25} = 4 Hence the intensity at this point is I = R'^{2} = {(\frac{R_{5}}{2})}^{2} = {(\frac{R_{5}}{R_{4}} \times \frac{R_{4}}{R_{3}} \times \frac{R_{3}}{R_{2}})}^{2} \times {(\frac{R_{2}}{2})}^{2} \begin{array}{r} or 1 = [0 .9]^{6} \\ I_{1} = 0 .531 I_{0} \end{array}$

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