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