MHT CET · Physics · Wave Optics
In a diffraction pattern due to single slit of width ' \(a\) ', the first minimum is observed at an angle \(30^{\circ}\) when light of wavelength \(5000 Å\) is incident on the slit. The first secondary maximum is observed at an angle \(\left[\sin 30=\frac{1}{2}\right]\)
- A \(\sin ^{-1}\left(\frac{1}{2}\right)\)
- B \(\sin ^{-1}\left(\frac{3}{4}\right)\)
- C \(\sin ^{-1}\left(\frac{1}{4}\right)\)
- D \(\sin ^{-1}\left(\frac{3}{5}\right)\)
Answer & Solution
Correct Answer
(B) \(\sin ^{-1}\left(\frac{3}{4}\right)\)
Step-by-step Solution
Detailed explanation
For \(\mathrm{n}^{\text {th }}\) secondary minimum, path difference \(=\mathrm{a} \sin \theta_{\mathrm{n}}=\mathrm{n} \lambda\) and for \(\mathrm{n}^{\text {th }}\) secondary maximum, path difference \(=a \sin \theta_n=(2 n+1) \frac{\lambda}{2}\)
\(\therefore \quad\) For \(1^{\text {st }}\) minimum, a \(\sin 30^{\circ}=\lambda\)...(i)
For \(2^{\text {nd }}\) maximum, a \(\sin \theta_n=(2+1) \frac{\lambda}{2}\)...(ii)
\(\therefore \quad\) Dividing equations (i) by equation (ii),
\(\frac{1 / 2}{\sin \theta_n}=\frac{2}{3} \Rightarrow \theta_n=\sin ^{-1}\left(\frac{3}{4}\right)\)
\(\therefore \quad\) For \(1^{\text {st }}\) minimum, a \(\sin 30^{\circ}=\lambda\)...(i)
For \(2^{\text {nd }}\) maximum, a \(\sin \theta_n=(2+1) \frac{\lambda}{2}\)...(ii)
\(\therefore \quad\) Dividing equations (i) by equation (ii),
\(\frac{1 / 2}{\sin \theta_n}=\frac{2}{3} \Rightarrow \theta_n=\sin ^{-1}\left(\frac{3}{4}\right)\)
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