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  • Angular width of central maximum in the Fraunhoffer's diffraction pattern is measured. Slit is illuminated by the light of another wavelength, angular width decrease by <img alt="30 \%" src="ht

Angular width of central maximum in the Fraunhoffer's diffraction pattern is measured. Slit is illuminated by the light of another wavelength, angular width decrease by 30 \%. Wavelength of light used is

Option: 1

3500 \AA


Option: 2

4200 \AA


Option: 3

4700 \AA


Option: 4

6000 \AA


Answers (1)

best_answer

For first diffraction min. d \sin \theta=\lambda
and if angle is small, \sin \theta=\theta

\mathrm{d} \theta=\lambda
i.e. Half angular width, \theta=\frac{\lambda}{d}
Full angular width w=2 \theta=\frac{2 \lambda}{d}

\text { Also } w^{\prime}=\frac{2 \lambda^{\prime}}{d}
\therefore \frac{\lambda^{\prime}}{\lambda}=\frac{w^{\prime}}{w} \text { or } \lambda^{\prime}=\lambda \frac{w^{\prime}}{w}
=6000 \times 0.7=4200 \AA
 

Posted by

Gautam harsolia

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