Q125 / 170 · Manipal (MET) 2018 · Physics

In a single slit diffraction pattern, the distance between the first minimum on the left and the first minimum on the right is 5 mm . The screen on which the diffraction pattern is displayed is at a distance of 80 cm from the slit. The wavelength is $6000 Å$.The slit width in ( mm ) is about

  • A
    0.576
  • B
    0.348
  • Correct answer
    C
    0.192
  • D
    0.096

Solution

Slit width, \(w=\frac{D \lambda}{d}\) \(\begin{aligned} & \text {Given, } D=80 \mathrm{~cm}=80 \times 10^{-2} \mathrm{~m} \\ & \lambda=6000 \AA=6000 \times 10^{-10} \mathrm{~m} \\ & d=\frac{5}{2} \mathrm{~mm}=\frac{5 \times 10^{-3}}{2} \mathrm{~m} \\ & \therefore w=\frac{80 \times 10^{-2} \times 6000 \times 10^{-10} \times 2}{5 \times 10^{-3}} \\ & w=0.192 \times 10^{-3} \mathrm{~m}=0.192 \mathrm{~mm} \end{aligned}\)

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

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