JEE Mains · Physics · STD 12 - 10. Wave optics
In single slit diffraction pattern, the wavelength of light used is \(628\) nm and slit width is \(0.2\) mm, the angular width of central maximum is \(\alpha \times 10^{-2}\) degrees. The value of \(\alpha\) is _______.
- A 32
- B 36
- C 38
- D 40
Answer & Solution
Correct Answer
(B) 36
Step-by-step Solution
Detailed explanation
The angular width of the central maximum in a single slit diffraction pattern is given by \(\Delta \theta = \dfrac{2\lambda}{a}\) Substituting the given values: \(\lambda = 628 \times 10^{-9}\) m \(a = 0.2 \times 10^{-3}\) m…
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