JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2025
- Medium
- MCQ
The Young's double slit interference experiment is performed using light consisting of 480 nm and 600 nm wavelengths to form interference patterns. The least number of the bright fringes of 480 nm light that are required for the first coincidence with the bright fringes formed by 600 nm light is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(5\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \frac{n_1 \lambda_1 D}{d}=\frac{n_2 \lambda_2 D}{d} \\
& n 480=m 600 \\
& n_{\min }=5
\end{aligned}\)
& \frac{n_1 \lambda_1 D}{d}=\frac{n_2 \lambda_2 D}{d} \\
& n 480=m 600 \\
& n_{\min }=5
\end{aligned}\)
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