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JEE Mains · Physics · STD 12 - Wave optics
  • JEE Mains 2020
  • Medium
  • Numerical

Visible light of wavelength \(6000 \times 10^{-8}\; \mathrm{cm}\) falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at \(60^{\circ}\) from the central maximum. If the first minimum is produced at \(\theta_{1},\) then \(\theta_{1}\) is close to.....\(^o\)

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Answer & Solution

Correct Answer

(D) \(25\)

Step-by-step Solution

Detailed explanation

\(\sin \theta=\frac{2 \lambda}{\omega}\)

\(\sin 60^{\circ}=\frac{2 \lambda}{\omega}\)

\(\sin \theta_{1}=\frac{\lambda}{\omega}=\frac{\sqrt{3}}{4}\)

\(\theta_{1}=25^{\circ}\)

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