Q142 / 180 · COMEDK 2014 · Physics

A screen is placed $2 \mathrm{~m}$ away from a narrow slit. It the first minimum lies 5 mm from either side of the central maxima, when plane waves of wavelength $5 \times 10^{-7} \mathrm{~m}$ are used, the slit width is

  • A
    $4.8 \times 10^{-4} \mathrm{~m}$
  • Correct answer
    B
    $2 \times 10^{-4} \mathrm{~m}$
  • C
    $5 \times 10^{-4} \mathrm{~m}$
  • D
    $24 \times 10^{-4} \mathrm{~m}$

Solution

Given, distance between source and screen, $D=2 \mathrm{~m}$ Distance of first minimum from the central maximum, $y=5 \mathrm{~mm}=5 \times 10^{-3} \mathrm{~m}$ Wavelength, $\lambda=5 \times 10^{-7} \mathrm{~m}$ We know that, $$ y_{n}=\frac{n \lambda D}{d} $$ For first minimum, $n=1$ $$ \begin{aligned} & & y_{1} &=\frac{\lambda D}{d} \\ \Rightarrow & & d &=\frac{\lambda D}{y_{1}}=\frac{5 \times 10^{-7} \times 2}{5 \times 10^{-3}}=2 \times 10^{-4} \mathrm{~m} \end{aligned} $$

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

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