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(D) The number of Half Period Zones $(HPZ)$ covered by a circular obstacle is given by the relation $n_1 b_1 = n_2 b_2$,where $n$ is the number of $HP

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$(\frac{R_9}{R_2})^2 I$

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(D) The number of Half Period Zones $(HPZ)$ covered by a circular obstacle is given by the relation $n_1 b_1 = n_2 b_2$,where $n$ is the number of $HPZ$ and $b$ is the distance from the disc.Given $n_1 = 1$ and $b_1 = 2 \, m = 200 \, cm$.For the second case,$b_2 = 25 \, cm$.Substituting the values: $1 	imes 200 = n_2 	imes 25 \implies n_2 = 8$.When $n$ zones are covered,the resultant amplitude $A$ is given by $A = A_1 - A_2 + A_3 - A_4 + ... \pm A_n$.For an even number of zones $n$,the resultant amplitude is $A = \frac{A_n}{2}$.Since intensity $I \propto A^2$,the intensity for $n=8$ is $I_2 = (A_8/2)^2$.Comparing this to the first case where $n=1$,$I_1 = A_1^2 = I$.Using the property of $HPZ$ amplitudes,$A_n \approx A_{n+1}$,the intensity at $n=8$ is equivalent to the amplitude contribution of the $9^{th}$ zone divided by $2$,i.e.,$I_2 = (R_9/2)^2$.Thus,$I_2 = (R_9/R_2)^2 I$.

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

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