MHT CET · Physics · Wave Optics
Light of wavelength \(5000 Å\) is incident normally on a slit. The first minimum of the diffraction pattern is observed to lie at distance of \(5 \mathrm{~mm}\) from the central maximum on a screen placed at a distance of \(2 \mathrm{~m}\) from the slit. The width of the slit.is
- A \(2 \mathrm{~cm}\)
- B \(0.2 \mathrm{~cm}\)
- C \(0.02 \mathrm{~cm}\)
- D \(0.01 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(C) \(0.02 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
The slit width is given as:
\(\mathrm{W}=\frac{\lambda \mathrm{D}}{\mathrm{d}}\)
\(\therefore \quad \mathrm{d}=\frac{2 \times 5000 \times 10^{-10}}{5 \times 10^{-3}}=0.02 \mathrm{~cm}\)
\(\mathrm{W}=\frac{\lambda \mathrm{D}}{\mathrm{d}}\)
\(\therefore \quad \mathrm{d}=\frac{2 \times 5000 \times 10^{-10}}{5 \times 10^{-3}}=0.02 \mathrm{~cm}\)
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