Wave Optics Question 6
Question 6 - 2024 (29 Jan Shift 1)
In a double slit experiment shown in figure, when light of wavelength $400 nm$ is used, dark fringe is observed at $P$. If $D=0.2 m$. the minimum distance between the slits $S _1$ and $S _2$ is $mm$.
Show Answer
Answer: (0.2)
Solution:
Path difference for minima at $P$
$$ \begin{aligned} & 2 \sqrt{D^{2}+d^{2}}-2 D=\frac{\lambda}{2} \text { } \\ & \therefore \sqrt{D^{2}+d^{2}}-D=\frac{\lambda}{4} \\ & \therefore \sqrt{D^{2}+d^{2}}=\frac{\lambda}{4}+D \\ & \Rightarrow D^{2}+d^{2}=D^{2}+\frac{\lambda^{2}}{16}+\frac{D \lambda}{2} \\ & \Rightarrow d^{2}=\frac{D \lambda}{2}+\frac{\lambda^{2}}{16} \\ & \Rightarrow d^{2}=\frac{0.2 \times 400 \times 10^{-9}}{2}+\frac{4 \times 10^{-14}}{4} \\ & \Rightarrow d^{2} \approx 400 \times 10^{-10} \\ & \therefore d=20 \times 10^{-5} \\ & \Rightarrow d=0.20 mm \end{aligned} $$
