The angular separation of the central maximum | vedclass

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(C) The angular width of the central maximum in a Fraunhofer diffraction pattern is given by the formula: $	heta = \frac{2 \lambda}{a}$,where $\lambd

Vedclass · Accepted Answer

$4800$

Vedclass · Answer

(C) The angular width of the central maximum in a Fraunhofer diffraction pattern is given by the formula: $	heta = \frac{2 \lambda}{a}$,where $\lambda$ is the wavelength of light and $a$ is the slit width.From this relation,we can see that $	heta \propto \lambda$.Let the initial wavelength be $\lambda_1 = 6000 Å$ and the initial angular width be $	heta_1$.When the wavelength is changed to $\lambda_2$,the angular width becomes $	heta_2 = 	heta_1 - 0.20 	heta_1 = 0.80 	heta_1$.Using the proportionality $\frac{	heta_2}{	heta_1} = \frac{\lambda_2}{\lambda_1}$,we get:$\frac{0.80 	heta_1}{	heta_1} = \frac{\lambda_2}{6000 Å}$$0.80 = \frac{\lambda_2}{6000 Å}$$\lambda_2 = 0.80 	imes 6000 Å = 4800 Å$.

The angular separation of the central maximum in the Fraunhofer diffraction pattern is measured. The slit is illuminated by light of wavelength $6000 Å$. If the slit is illuminated by light of another wavelength,the angular separation decreases by $20 \%$. The wavelength of light used is (in $Å$)

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