Q87 / 180 · AIIMS 2018 (26 May) · Physics

In a single slit diffraction the distance between plane of slit and screen is $1 \mathrm{~m}$. The size of the slit is $0.7 \mathrm{~mm}$ and second maximum is formed at the distance of $2 \mathrm{~mm}$ from the centre of the screen, then find out the wavelength of light.

  • Correct answer
    A
    $5600 Å$
  • B
    $2800 Å$
  • C
    $3000 Å$
  • D
    $6000 Å$

Solution

For second maxima path difference, $\begin{aligned} & a \tan \theta=\frac{5 \lambda}{2} ; a \cdot \frac{y}{D}=\frac{5 \lambda}{2} \\ & \lambda=\frac{2 a y}{5 D}=\frac{2 \times 0.7 \times 10^{-3} \times 2 \times 10^{-3}}{5 \times 1} \\ & =\frac{2.8}{5} \times 10^{-6} \mathrm{~m}=\frac{28}{5} \times 10^{-7}=5600 Å \\ & \end{aligned}$

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

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