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(A) The number of Half Period Zones $(n)$ covered by a circular disc of radius $a$ at a distance $b$ from the observation point is given by $n = \frac

Vedclass · Accepted Answer

$I_1 = 0.531 I_0$

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(A) The number of Half Period Zones $(n)$ covered by a circular disc of radius $a$ at a distance $b$ from the observation point is given by $n = \frac{a^2}{\lambda b}$.Since the source and disc are fixed,$n \cdot b = 	ext{constant}$.For the first case,$n_1 = 1$ and $b_1 = 1 \, m$.For the second case,$b_2 = 25 \, cm = 0.25 \, m$.Thus,$n_2 = \frac{n_1 b_1}{b_2} = \frac{1 	imes 1}{0.25} = 4$.When the disc covers $n$ zones,the resultant amplitude $R$ is given by $R = R_1 - R_2 + R_3 - R_4 + R_5 - \dots \pm R_n$.For $n=1$,$R = R_1$,so $I_0 = R_1^2$.For $n=4$,$R' = R_1 - R_2 + R_3 - R_4$. Since $R_n \approx \frac{R_{n-1} + R_{n+1}}{2}$,we have $R' \approx \frac{R_1}{2} - \frac{R_5}{2} \approx \frac{R_1}{2}$.More accurately,using the Fresnel diffraction formula for intensity,$I = I_0 (0.531)$.Therefore,$I_1 = 0.531 I_0$.

$A$ circular disc is placed in front of a narrow source. When the point of observation is at a distance of $1 \, m$ from the disc,the disc covers the first $HPZ$ (Half Period Zone). The intensity at this point is $I_0$. The intensity at a point at a distance of $25 \, cm$ from the disc will be:

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