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  • If light with wavelength falls on a slit of width <img alt="\mathrm{10 \mathrm{~mm}

If light with wavelength 0.50 \mathrm{~mm} falls on a slit of width \mathrm{10 \mathrm{~mm}} and at an angle \mathrm{\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

\mathrm{\text { None of these }}


Answers (1)

best_answer

For first diffraction minima at angle \mathrm{\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}

Posted by

Divya Prakash Singh

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