JEE Mains · Physics · STD 12 - 10. Wave optics
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 \(..............\,mm\)
- A \(4\)
- B \(1.33\)
- C \(3\)
- D \(2\)
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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