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Light of wavelength 589.3 \mathrm{~nm} is incident normally on two slits of separation 0.1 \mathrm{~mm} width 0.01 \mathrm{~nm}. The angular width of the central diffraction maximum at a distance of 1 \mathrm{~m} from the slits, is:

Option: 1

0.68^{\circ}


Option: 2

-0.34^{\circ}


Option: 3

2.05^{\circ}


Option: 4

\mathrm{None \, \, of \, \, these}


Answers (1)

best_answer

\mathrm{ \beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{589.3 \times 10^{-9}}{0.1 \times 10^{-3}} \mathrm{D} }

Hence, angular width of the central diffraction.

\mathrm{\begin{aligned} & \theta=\frac{\beta}{D}=\frac{589.3 \times 10^{-9}}{0.1 \times 10^{-3}} \mathrm{rad} \\\\ & =589.3 \times 10^{-5} \times \frac{180}{\pi}=0.34^{\circ} \end{aligned}}

Posted by

Ritika Kankaria

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