Q121 / 150 · MHT CET 2024 (03 May Shift 1) · Physics

A single slit diffraction pattern is formed with light of wavelength $6195 Å$. The second secondary maximum for this wavelength coincides with the third secondary maximum in the pattern for light of wavelength ' $\lambda_0$ '. The value of ' $\lambda_0$ ' is

  • A
    $4180 Å$
  • Correct answer
    B
    $4425 Å$
  • C
    $5330 Å$
  • D
    $6235 Å$

Solution

For $\mathrm{n}^{\text {th }}$ secondary maximum, $\begin{array}{ll} & \mathrm{x}_{\mathrm{n}}=\frac{(2 \mathrm{n}+1) \lambda \mathrm{D}}{2 \mathrm{a}} \\ \therefore \quad & \text { For } \mathrm{n}=2, \\ & \mathrm{x}_2=\frac{5 \lambda \mathrm{D}}{2 \mathrm{a}} \\ & \text { and for } \mathrm{n}=3, \mathrm{x}_3=\frac{7 \lambda_0 \mathrm{D}}{2 \mathrm{a}} \\ & \text { Here, } \mathrm{x}_2=\mathrm{x}_3 \\ \therefore \quad & 5 \lambda=7 \lambda_0 \\ \therefore \quad & \lambda_0=\frac{5 \lambda}{7}=\frac{5 \times 6195}{7}=4425 Å \end{array}$

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

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