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JEE Mains · Physics · STD 12 - Wave optics
  • JEE Mains 2023
  • Medium
  • Numerical

The width of fringe is \(2\,mm\) on the screen in a double slits experiment for the light of wavelength of \(400\,nm\). The width of the fringe for the light of wavelength \(600\,nm\) will be \(..............\,mm\)

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Answer & Solution

Correct Answer

(C) \(3\)

Step-by-step Solution

Detailed explanation

\(\text { Fringe width }(\beta)=\frac{ D \lambda}{ d }\)

\(\Rightarrow \frac{\beta_2}{\beta_1}=\frac{\lambda_2}{\lambda_1}\)

\(\Rightarrow \frac{\beta_2}{2\,mm }=\frac{600\,nm }{400\,nm }=\frac{3}{2}\)

\(\Rightarrow \beta_2=3\,mm\)

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