AP EAMCET · PHYSICS · Wave Optics
In Young's double slit experiment the slits are \(3 \mathrm{~mm}\) apart and are illuminated by light of two wavelengths \(3750 Å\) and \(7500 Å\). The screen is placed at \(4 \mathrm{~m}\) from the slits. The minimum distance from the common central bright fringe on the screen at which the bright fringe of one interference pattern due to one wavelength coincide with the bright fringe of the other is
- A \(2 \mathrm{~mm}\)
- B \(3 \mathrm{~mm}\)
- C \(1 \mathrm{~mm}\)
- D \(8 \mathrm{~mm}\)
Answer & Solution
Correct Answer
(C) \(1 \mathrm{~mm}\)
Step-by-step Solution
Detailed explanation
Given, \(\lambda_1=3750 Å\) \(\begin{aligned} \lambda_2 & =7500 Å \\ D & =4 \mathrm{~m} \\ d & =3 \mathrm{~mm}=3 \times 10^{-3} \mathrm{~m}\end{aligned}\) since, \(\frac{\lambda_1}{\lambda_2}=\frac{3750}{7500}=\frac{1}{2}\) We know that,…
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