JEE PHYSICS MCQ: WAVE OPTICS

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In young’s single slit experiment with sodium vapour lamp of wavelength 589 nm and the slits 0.589 mm apart, the half angular width of the central maximum is

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Correct option: C

{\sin ^{ - 1}}\left[ {0.001} \right]

Option C (

{\sin ^{ - 1}}\left[ {0.001} \right]

) is correct because the half angular width of the central maximum in single-slit diffraction is given by

\sin \theta = \frac{\lambda}{d}

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