Q133 / 150 · MHT CET 2024 (03 May Shift 2) · Physics

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
    $\frac{\mathrm{d}^2}{2 \mathrm{D}}$
  • Correct answer
    C
    $\frac{\mathrm{d}^2}{3 \mathrm{D}}$
  • D
    $\frac{\mathrm{d}^2}{4 \mathrm{D}}$

Solution

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 $\mathrm{n}=2$ $\therefore \quad \frac{\mathrm{d}}{2}=\frac{(2 \times 2-1) \lambda \mathrm{D}}{2 \mathrm{~d}}$ $\therefore \quad \lambda=\frac{\mathrm{d}^2}{3 \mathrm{D}}$

Hence, the correct answer is option: 'C'.

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