In a Fraunhofer diffraction due to a single s | vedclass

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(A) For Fraunhofer diffraction at a single slit,the position of the $n^{th}$ minimum is given by the formula: $x_n = \frac{n f \lambda}{a}$.Given: $n

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$5000$

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(A) For Fraunhofer diffraction at a single slit,the position of the $n^{th}$ minimum is given by the formula: $x_n = \frac{n f \lambda}{a}$.Given: $n = 3$,$f = 1 \ m$,$a = 0.3 \ mm = 3 	imes 10^{-4} \ m$,and $x_n = 5 \ mm = 5 	imes 10^{-3} \ m$.Rearranging the formula for wavelength $\lambda$: $\lambda = \frac{a x_n}{n f}$.Substituting the values: $\lambda = \frac{(3 	imes 10^{-4} \ m) 	imes (5 	imes 10^{-3} \ m)}{3 	imes 1 \ m}$.$\lambda = 5 	imes 10^{-7} \ m$.Converting to $\mathring{A}$: $\lambda = 5 	imes 10^{-7} 	imes 10^{10} \ \mathring{A} = 5000 \ \mathring{A}$.

Similar Questions

In a single-slit diffraction experiment,the slit is illuminated by light of two wavelengths $\lambda_1$ and $\lambda_2$. It is observed that the $2^{nd}$ order diffraction minimum for $\lambda_1$ coincides with the $3^{rd}$ diffraction minimum for $\lambda_2$. Then:

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