JEE Mains · Physics · STD 12 - 10. Wave optics
In a Young's double slit experiment, an angular width of the fringe is \(0.35^{\circ}\) on a screen placed at \(2\,m\) away for particular wavelength of \(450\,nm\). The angular width of the fringe, when whole system is immersed in a medium of refractive index \(7 / 5\), is \(\frac{1}{\alpha}\). The value of \(\alpha\) is ..............
- A \(1\)
- B \(4\)
- C \(5\)
- D \(0\)
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
\(\beta=\frac{0.35 \times 5}{7}=0.25\) \(\frac{1}{\alpha}=\frac{2.5}{100}\) \(\alpha=4\)
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