JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2024
- Hard
- Numerical
In Young's double slit experiment, carried out with light of wavelength \(5000\) \(A\), the distance between the slits is \(0.3 \mathrm{~mm}\) and the screen is at \(200 \mathrm{~cm}\) from the slits. The central maximum is at \(x=0 \mathrm{~cm}\). The value of \(\mathrm{x}\) for third maxima is _______ \(\mathrm{mm}\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(10\)
Step-by-step Solution
Detailed explanation
\(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{5 \times 10^{-7} \times 2}{3 \times 10^{-4}}=\frac{10 \times 10^{-3}}{3} \mathrm{~m}\)
For \(3^{\text {rd }}\) maxima \(y_3=3 \beta=10 \times 10^{-3} \mathrm{~m}=10 \mathrm{~mm}\)
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