MHT CET · Physics · Wave Optics
In Fraunhofer diffraction pattern, slit width is \(0.2 \mathrm{~mm}\) and screen is at \(2 \mathrm{~m}\) away from the lens. If wavelength of light used is \(5000 Å\) then the distance between the first minimum on either side of the central maximum is \((\theta\) is small and measured in radian)
- A \(2 \times 10^{-2} \mathrm{~m}\)
- B \(10^{-1} \mathrm{~m}\)
- C \(10^{-2} \mathrm{~m}\)
- D \(10^{-3} \mathrm{~m}\)
Answer & Solution
Correct Answer
(C) \(10^{-2} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
Distance between the first minima on either side of the central maximum \(=\frac{2 \lambda \mathrm{D}}{\mathrm{a}}=\frac{2 \times 5 \times 10^{-7} \times 2}{0.2 \times 10^{-3}}=10^{-2} \mathrm{~m}\)
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