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Wave Optics question

2024 · 27 Jan · Shift 2 · Q88
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Wave Optics question

2024 · 27 Jan · Shift 2 · Q88

JEE MainPhysicsWave OpticsNumerical+4 / −1
A parallel beam of monochromatic light of wavelength 5000 Ao\mathop A\limits^oAo​ is incident normally on a single narrow slit of width 0.001 mm0.001 \mathrm{~mm}0.001 mm. The light is focused by convex lens on screen, placed on its focal plane. The first minima will be formed for the angle of diffraction of ‾\underline{\hspace{2cm}}​ (degree).
Numerical answer
View written solutionFree

Correct answer: 30

  1. Given data
  • Wavelength: λ=5000 A˚=5000×10−10 m=5×10−7 m\lambda = 5000\,\text{Å} = 5000\times 10^{-10}\,\text{m} = 5\times 10^{-7}\,\text{m}λ=5000A˚=5000×10−10m=5×10−7m
  • Slit width: a=0.001 mm=10−3×10−3 m=10−6 ma = 0.001\,\text{mm} = 10^{-3}\times 10^{-3}\,\text{m} = 10^{-6}\,\text{m}a=0.001mm=10−3×10−3m=10−6m
  1. Condition for first minimum in single-slit diffraction

For a single slit, minima occur at asin⁡θ=mλa\sin\theta = m\lambdaasinθ=mλ where m=1,2,3,…m=1,2,3,\dotsm=1,2,3,…

For the first minimum, m=1m=1m=1: asin⁡θ=λa\sin\theta = \lambdaasinθ=λ

  1. Substitute the values

sin⁡θ=λa=5×10−710−6=0.5\sin\theta = \frac{\lambda}{a} = \frac{5\times 10^{-7}}{10^{-6}} = 0.5sinθ=aλ​=10−65×10−7​=0.5

  1. Find the angle

θ=sin⁡−1(0.5)=30∘\theta = \sin^{-1}(0.5) = 30^\circθ=sin−1(0.5)=30∘

  1. Final answer

The first minimum is formed at diffraction angle 30∘\boxed{30^\circ}30∘​

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