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If light with wavelength 0.50 \mathrm{~mm} falls on a slit of width 10 \mathrm{~mm} and at an angle \theta^{\circ}=30^{\circ} to its normal. Then angular position of first minima located on right sides of the central Fraunhoffer's diffraction will be at

Option: 1

33.4^{\circ}


Option: 2

26.8^{\circ}


Option: 3

39.8^{\circ}


Option: 4

None of these


Answers (1)

best_answer

For first diffraction minima at angle \theta
\mathrm{d\left(\sin \theta-\sin \theta_{0}\right)= \pm \lambda}

For right of C.M.,
\mathrm{\sin \theta_{1}=\sin \theta_{0}+\frac{\lambda}{d}}
\mathrm{=0.5+\frac{0.5}{10}=0.55\, i.e., \theta_{1}=33.37^{0}}

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