The angular width of the central maxima in Fr | vedclass

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Question

(B) The angular width of the central maxima in Fraunhofer diffraction is given by $	heta = \frac{2\lambda}{d}$,where $\lambda$ is the wavelength and

Vedclass · Accepted Answer

$4200$

Vedclass · Answer

(B) The angular width of the central maxima in Fraunhofer diffraction is given by $	heta = \frac{2\lambda}{d}$,where $\lambda$ is the wavelength and $d$ is the slit width.Since $	heta \propto \lambda$,we have the ratio $\frac{	heta_2}{	heta_1} = \frac{\lambda_2}{\lambda_1}$.Given that the angular width decreases by $30 \%$,the new angular width $	heta_2 = 	heta_0 - 0.30	heta_0 = 0.70	heta_0$.Substituting the values: $\frac{0.70	heta_0}{	heta_0} = \frac{\lambda_2}{6000 \; \mathring{A}}$.$0.70 = \frac{\lambda_2}{6000 \; \mathring{A}}$.$\lambda_2 = 0.70 	imes 6000 \; \mathring{A} = 4200 \; \mathring{A}$.

The angular width of the central maxima in Fraunhofer diffraction for $\lambda = 6000 \; \mathring{A}$ is $\theta_{0}$. When the same slit is illuminated by another monochromatic light,the angular width decreases by $30 \%$. The wavelength of this light is ....... $\mathring{A}$.

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