Q167 / 180 · COMEDK 2017 · Physics

A single slit of width $a$ is illuminated by violet light of wavelength $400 \mathrm{~nm}$ and the width of the diffraction pattern is measured as $y$. When half of the slit width is covered and illuminated by yellow light of wavelength $600 \mathrm{~nm}$, the width of the diffraction pattern is

  • A
    the pattern vanishes and the width is zero
  • B
    $\frac{y}{3}$
  • Correct answer
    C
    $3 y$
  • D
    None of the above

Solution

Given, In first case, Wavelength of violet light, $\lambda_{1}=400 \mathrm{~nm}$ We know that, width of the diffraction pattern is given as $\begin{aligned} & \beta=y=\frac{2 D \lambda_{1}}{d} \\ \Rightarrow \quad & y=\frac{2 D \lambda_{1}}{d}=\frac{2 D \times 400}{d} \end{aligned}$ $y=\frac{800 D}{d}...(i)$ In the second case, $\begin{aligned} \lambda_{2} &=600 \mathrm{~nm}, d^{\prime}=\frac{d}{2} \\ \therefore \quad y^{\prime} &=\frac{2 D \lambda^{\prime}}{d^{\prime}}=\frac{2 D \times 600}{\frac{d}{2}} \\ &=2400 \frac{D}{d}=3 \times \frac{800 D}{d} \quad \text { [from Eq. (i)] } \\ &=3 y \end{aligned}$

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

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