Question

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 The intensity at a point distance 25 cm from the disc will be (If ratio of consecutive amplitude of HPZ is 0.9)

Answer 1

(0.9)^6 I_1

Answer 2

1. The number of HPZs covered by a disc at a distance $b$ from the observation point is proportional to $1/b$ when the source is far away ($a \gg b$). Given the disc covers 1 HPZ at $b=1$ m, it covers $N_2 = (1/0.25) \times 1 = 4$ HPZs at $b=0.25$ m.

2. When a disc covers the first $N$ HPZs, the amplitude at the observation point is approximately half the amplitude contribution of the $(N+1)$-th HPZ, i.e., $A \approx \frac{1}{2} A_{N+1}$.

3. At $b=1$ m, $N=1$, so the amplitude $A_1 \approx \frac{1}{2} A_2$. The intensity $I_1 \propto A_1^2 \propto (\frac{1}{2} A_2)^2$.

4. At $b=0.25$ m, $N=4$, so the amplitude $A_2 \approx \frac{1}{2} A_5$. The intensity $I_2 \propto A_2^2 \propto (\frac{1}{2} A_5)^2$.

5. The ratio of consecutive amplitudes is $k=0.9$. Thus, $A_5 = k^{5-2} A_2 = (0.9)^3 A_2$.

6. Substitute $A_5$ in the expression for $I_2$: $I_2 \propto (\frac{1}{2} (0.9)^3 A_2)^2 = (0.9)^6 (\frac{1}{2} A_2)^2$.

7. Comparing $I_1$ and $I_2$, we get $I_2 = (0.9)^6 I_1$.