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QJEEM20Ph5b4d: JEE Main 2020 Physics - Class 12 JEEM20Ph5b4d Solution: Visible light of wavelength 6000 10 -8 cm falls normally on a…

Visible light of wavelength 6000 ×\times 10-8 cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60o from the central maximum. If the first minimum is produced at θ\theta 1, then θ\theta 1, is close to :

Multiple Choice Options

  • A 45o
  • B 30o
  • C 25o
  • D 20o

Step-by-Step Verified Solution

For 2nd minima

sin⁡θ=2λd\sin \theta = {{2\lambda } \over d }

⇒\Rightarrow sin⁡60∘=2λd\sin 60^\circ = {{2\lambda } \over d}

⇒\Rightarrow λd=34{\lambda \over d} = {{\sqrt 3 } \over 4}

For 1st minima

sin⁡θ1=λd\sin {\theta _1} = {\lambda \over d} = 34{{\sqrt 3 } \over 4}

⇒\Rightarrow θ1={\theta _1} = 25o

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