MHT CET · Physics · Wave Optics
Light of wavelength ' \(\lambda\) ' is incident on a single slit of width ' \(a\) ' and distance between slit and screen is ' \(D\) '. In diffraction pattern, if slit width is equal to the width of the central maximum then \(\mathrm{D}=\)
- A \(\frac{\mathrm{a}^2}{\lambda}\)
- B \(\frac{\mathrm{a}}{\lambda}\)
- C \(\frac{a^2}{2 \lambda}\)
- D \(\frac{\mathrm{a}}{2 \lambda}\)
Answer & Solution
Correct Answer
(C) \(\frac{a^2}{2 \lambda}\)
Step-by-step Solution
Detailed explanation
Width of central maximum \(=\frac{2 \lambda \mathrm{D}}{\mathrm{a}}=\mathrm{a}\)
\(
\therefore \mathrm{D}=\frac{\mathrm{a}^2}{2 \lambda}
\)
\(
\therefore \mathrm{D}=\frac{\mathrm{a}^2}{2 \lambda}
\)
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