In a Fraunhofer diffraction of a single slit, | vedclass

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

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$4860$

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(C) The angular width of the central maximum in a single-slit Fraunhofer diffraction is given by $	heta = \frac{2\lambda}{a}$, where $\lambda$ is the wavelength of light and $a$ is the slit width.Let the initial wavelength be $\lambda_1 = 6480 	ext{ Å}$ and the initial angular width be $	heta_1 = \frac{2\lambda_1}{a}$.When the wavelength is changed to $\lambda_2 = \lambda$, the new angular width is $	heta_2 = \frac{2\lambda}{a}$.According to the problem, the angular width decreases by $25\%$, so $	heta_2 = 	heta_1 - 0.25	heta_1 = 0.75	heta_1$.Substituting the expressions for $	heta_1$ and $	heta_2$:$\frac{2\lambda}{a} = 0.75 	imes \frac{2\lambda_1}{a}$.This simplifies to $\lambda = 0.75 	imes \lambda_1$.$\lambda = 0.75 	imes 6480 	ext{ Å} = 4860 	ext{ Å}$.

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