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  • Angular width of central maximum in the Fraunhofer's diffraction pattern is measure. Slit is illuminated by the light of wavelength <img alt="\mathrm{6000 \, \AA}" src="https://entrancecorner.o

Angular width of central maximum in the Fraunhofer's diffraction pattern is measure. Slit is illuminated by the light of wavelength \mathrm{6000 \, \AA}. If slit is illuminated by light of another wavelength, angular width decreases by \mathrm{30 \%}. Wavelength of light used is:

Option: 1

\mathrm{3500\, \, \AA}


Option: 2

\mathrm{4200 \, \AA}


Option: 3

\mathrm{4700\, \AA}


Option: 4

\mathrm{6000\, \AA}


Answers (1)

best_answer

In case of single slit diffraction experiment the angular width of central maximum is given by

\mathrm{ \theta=\frac{2 \lambda}{\mathrm{a}} }

Where, \mathrm{\mathrm{a}=} width of the slit

\mathrm{\lambda = } wavelength of light

\mathrm{ \therefore \frac{\theta^{\prime}}{\theta}=\frac{\lambda^{\prime}}{\lambda} }

According to given problem
\mathrm{ \begin{aligned} & \theta^{\prime}=\theta-\frac{30}{100} \theta=0.70 \theta \\\\ & \therefore \quad \frac{\lambda^{\prime}}{\lambda}=\frac{0.7 \theta}{\theta}=0.7 \\\\ & \lambda^{\prime}=0.7 \lambda=0.7 \times 6000 \AA=4200 \AA \end{aligned} }

Posted by

Ritika Jonwal

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