MHT CET · Physics · Wave Optics
In Young's double slit experiment, in an interference pattern, second minimum is observed exactly in front of one slit. The distance between the two coherent sources is ' \(d\) ' and the distance between the source and screen is ' \(D\) '. The wave length of light \((\lambda)\) used is
- A \(\frac{\mathrm{d}^2}{\mathrm{D}}\)
- B \(\frac{\mathrm{d}^2}{2 \mathrm{D}}\)
- C \(\frac{\mathrm{d}^2}{3 \mathrm{D}}\)
- D \(\frac{\mathrm{d}^2}{4 \mathrm{D}}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{d}^2}{3 \mathrm{D}}\)
Step-by-step Solution
Detailed explanation
Second minimum is exactly in front of one slit indicates, \(\mathrm{y}_2=\frac{\mathrm{d}}{2}\)
But \(\mathrm{y}_{\mathrm{n}}^{\prime}=\frac{(2 \mathrm{n}-1) \lambda \mathrm{D}}{2 \mathrm{~d}}\)
For \(\mathrm{n}=2\)
\(\therefore \quad \frac{\mathrm{d}}{2}=\frac{(2 \times 2-1) \lambda \mathrm{D}}{2 \mathrm{~d}}\)
\(\therefore \quad \lambda=\frac{\mathrm{d}^2}{3 \mathrm{D}}\)
But \(\mathrm{y}_{\mathrm{n}}^{\prime}=\frac{(2 \mathrm{n}-1) \lambda \mathrm{D}}{2 \mathrm{~d}}\)
For \(\mathrm{n}=2\)
\(\therefore \quad \frac{\mathrm{d}}{2}=\frac{(2 \times 2-1) \lambda \mathrm{D}}{2 \mathrm{~d}}\)
\(\therefore \quad \lambda=\frac{\mathrm{d}^2}{3 \mathrm{D}}\)
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