JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2025
- Hard
- Numerical
A double slit interference experiment performed with a light of wavelength 600 nm forms an interference fringe pattern on a screen with 10 th bright fringe having its centre at a distance of 10 mm from the central maximum. Distance of the centre of the same 10th bright fringe from the central maximum when the source of light is replaced by another source of wavelength 660 nm would be _________ mm .
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 11
Step-by-step Solution
Detailed explanation
Position of the \(n^{\text {th }}\) bright fringe w.r.t. central maxima in a YDSE is \(y_n=n \frac{\lambda D}{d}\).
\(\Rightarrow \quad \frac{y_{10}^{\prime}}{y_{10}}=\frac{\lambda^{\prime}}{\lambda}\)
or \(\quad y_{10}^{\prime}=\frac{\lambda^{\prime}}{\lambda} y_{10}\)
\(\begin{aligned} & =\left(\frac{660 \mathrm{~nm}}{600 \mathrm{~nm}}\right) 10 \mathrm{~mm} \\ & =11 \mathrm{~mm}\end{aligned}\)
\(\Rightarrow \quad \frac{y_{10}^{\prime}}{y_{10}}=\frac{\lambda^{\prime}}{\lambda}\)
or \(\quad y_{10}^{\prime}=\frac{\lambda^{\prime}}{\lambda} y_{10}\)
\(\begin{aligned} & =\left(\frac{660 \mathrm{~nm}}{600 \mathrm{~nm}}\right) 10 \mathrm{~mm} \\ & =11 \mathrm{~mm}\end{aligned}\)
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