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JEE Mains · Physics · STD 12 - Wave optics
  • JEE Mains 2021
  • Hard
  • Numerical

In a Young's double slit experiment, the slits are separated by \(0.3\, {mm}\) and the screen is \(1.5\, {m}\) away from the plane of slits. Distance between fourth bright fringes on both sides of central bright is \(2.4\, {cm}\). The frequency of light used is \(..........\,\times 10^{14} {Hz}\)

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Answer & Solution

Correct Answer

(D) \(5\)

Step-by-step Solution

Detailed explanation

\(8 \beta=2.4 \,{cm}\)

\(\frac{8 \lambda \Delta}{{d}}=2.4\, {cm}\)

\(\frac{8 \times 1.5 \times {c}}{0.3 \times 10^{-5} \times {f}}=2.4 \times 10^{-2}\)

\({f}=5 \times 10^{14}\, {Hz}\)

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