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 \(..............\,mm\)
Pick an option to check your answer.
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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