In a double slit experiment shown in figure, when light of wavelength $400\ \text{nm}$ is used, a dark fringe is observed at $P$. If $D = 0.2\ \text{m}$, the minimum distance between the slits $S_1$ and $S_2$ is $\alpha\ \text{mm}$. Write the value of $10\alpha$ to the nearest integer.
Numerical value type. Enter your answer.
Answer: 2
$S_2$ lies on the line joining the source and $P$, so the path via $S_2$ is $2D$. The path via $S_1$ is $2\sqrt{D^2 + d^2}$.
For a dark fringe at P with the smallest $d$:
$$2\left(\sqrt{D^2 + d^2} - D\right) = \frac\lambda2$$
Since $d \ll D$, $\sqrt{D^2 + d^2} - D \approx \dfrac{d^2}{2D}$:
$$\frac{d^2}{D} = \frac\lambda2 \;\Rightarrow\; d^2 = \frac{0.2\times400\times10^{-9}}{2} = 4\times10^{-8}\ \text{m}^2$$
$$d = 2\times10^{-4}\ \text{m} = 0.2\ \text{mm} \;\Rightarrow\; 10\alpha = 2$$
Solution by Sreeraj P, M.Sc Physics