Question

On the '+y' axis, two identical coherent sources $S_1$ and $S_2$, separated by distance 'd' emits light wave each of wave length '$\lambda$' = 4000nm. Plane x=0 serves as an imaginary boundary between two medium of refractive index $\mu$ = 1 and $\mu$ = 1.125. If a maxima is observed just on the left of origin and a minima is observe just on the right of origin, then the minimum non-zero value of d (in $\mu m$) is 2x. Then x is

• A. 32
• B. 64
• C. 128
• D. 256

Answer 1

Answer: (B)

64

Answer 2

The path difference for constructive interference in the first medium ($\mu_1=1$) is $d = n\lambda$, where $n$ is an integer. The path difference for destructive interference in the second medium ($\mu_2=1.125$) is $\mu_2 d = (m + 1/2)\lambda'$, where $\lambda'$ is the wavelength in the second medium. The wavelength in the second medium is $\lambda' = \lambda/\mu_2$. Thus, $\mu_2 d = (m + 1/2)(\lambda/\mu_2)$. Substituting $d=n\lambda$ into the second equation gives $\mu_2 (n\lambda) = (m + 1/2)(\lambda/\mu_2)$. This simplifies to $\mu_2^2 n = m + 1/2$. Substituting the values, $(1.125)^2 n = m + 0.5$, which is $(9/8)^2 n = m + 1/2$, or $81/64 n = m + 1/2$. Multiplying by 64, we get $81n = 64m + 32$, or $81n - 64m = 32$. This is a linear Diophantine equation. Using the Extended Euclidean Algorithm, we find that $81(480) - 64(608) = 32$. The general solution is $n = 480 - 64k$ and $m = 608 - 81k$. We need the minimum positive integer value for $n$. For $n>0$, $480 - 64k > 0$, so $k < 480/64 = 7.5$. The largest integer value for $k$ is 7. When $k=7$, $n = 480 - 64(7) = 480 - 448 = 32$. This is the minimum positive integer value for $n$. The minimum non-zero value of $d$ is $d = n\lambda = 32 \times 4000$ nm $= 128000$ nm. Converting to $\mu$m, $d = 128 \mu$m. The problem states $d = 2x$, so $2x = 128 \mu$m, which means $x = 64$.