KCET · Physics · Wave Optics
Light of wavelength \( 600 \mathrm{~nm} \) is incident normally on a slit of width \( 0.2 \mathrm{~mm} \). The angular width of
central maxima in the diffraction pattern is (measured from minimum to minimum)
- A \( 6 \times 10^{-3} \mathrm{rad} \)
- B \( 4 \times 10^{-3} \mathrm{rad} \)
- C \( 2.4 \times 10^{-3} \mathrm{rad} \)
- D \( 4.5 \times 10^{-3} \mathrm{rad} \)
Answer & Solution
Correct Answer
(A) \( 6 \times 10^{-3} \mathrm{rad} \)
Step-by-step Solution
Detailed explanation
Given, wavelength, \(\lambda=600 \mathrm{~nm}=600 \times 10^{-9} \mathrm{~m}\) width of slit, \(a=0.2 \mathrm{~mm}=0.2 \times 10^{-3} \mathrm{~m}\)
Angular width of central maxima \(=\frac{2 \lambda}{a}=\frac{2 \times 600 \times 10^{-9}}{0.2 \times 10^{-3}}\)
\(=600 \times 10^{-5}=6 \times 10^{-3}\)
Therefore, angular width of central maxima \(=6 \times 10^{-3} \mathrm{rad}\)
Angular width of central maxima \(=\frac{2 \lambda}{a}=\frac{2 \times 600 \times 10^{-9}}{0.2 \times 10^{-3}}\)
\(=600 \times 10^{-5}=6 \times 10^{-3}\)
Therefore, angular width of central maxima \(=6 \times 10^{-3} \mathrm{rad}\)
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