MHT CET · Physics · Wave Optics
In Young's double slit experiment, the two slits are 'd' distance apart. Interference pattern is observed on a screen at a distance ' \(D\) ' from the slits. A dark fringe is observed on a screen directly opposite to one of the slits. The wavelength of light is
- A \(\frac{D^2}{2 d}\)
- B \(\frac{\mathrm{d}^2}{2 \mathrm{D}}\)
- C \(\frac{D^2}{d}\)
- D \(\frac{\mathrm{d}^2}{\mathrm{D}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{d}^2}{\mathrm{D}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} S_2 P & =\left(D^2+d^2\right)^{1 / 2} \\ & =D\left[1+\frac{d^2}{D^2}\right]^{1 / 2}\end{aligned}\)
Using binomial equation,
\(\mathrm{S}_2 \mathrm{P}=\mathrm{D}\left[1+\frac{1}{2} \frac{\mathrm{d}^2}{\mathrm{D}^2}\right]^{1 / 2}=\mathrm{D}+\frac{\mathrm{d}^2}{2 \mathrm{D}}\)
\(\Rightarrow\) Path difference \(=\frac{d^2}{2 D}\)
For dark fringe, \(\frac{\mathrm{d}^2}{2 \mathrm{D}}=\frac{\lambda}{2}\)
\(\therefore \quad \lambda=\frac{\mathrm{d}^2}{\mathrm{D}}\)
Using binomial equation,
\(\mathrm{S}_2 \mathrm{P}=\mathrm{D}\left[1+\frac{1}{2} \frac{\mathrm{d}^2}{\mathrm{D}^2}\right]^{1 / 2}=\mathrm{D}+\frac{\mathrm{d}^2}{2 \mathrm{D}}\)
\(\Rightarrow\) Path difference \(=\frac{d^2}{2 D}\)
For dark fringe, \(\frac{\mathrm{d}^2}{2 \mathrm{D}}=\frac{\lambda}{2}\)
\(\therefore \quad \lambda=\frac{\mathrm{d}^2}{\mathrm{D}}\)
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