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

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$5 	imes 10^{-7} \ m$

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(C) The angular width of the central maximum in a single slit diffraction pattern is given by $	heta = \frac{2\lambda}{d}$,where $\lambda$ is the wavelength and $d$ is the slit width.The change in angular width is $\Delta	heta = \frac{2\Delta\lambda}{d}$,where $\Delta\lambda = |\lambda_1 - \lambda_2|$.Given frequencies $f_1 = 4 	imes 10^{14} \ s^{-1}$ and $f_2 = 5 	imes 10^{14} \ s^{-1}$.Using $\lambda = \frac{c}{f}$,where $c = 3 	imes 10^8 \ m/s$:$\lambda_1 = \frac{3 	imes 10^8}{4 	imes 10^{14}} = 0.75 	imes 10^{-6} \ m = 7.5 	imes 10^{-7} \ m$.$\lambda_2 = \frac{3 	imes 10^8}{5 	imes 10^{14}} = 0.60 	imes 10^{-6} \ m = 6.0 	imes 10^{-7} \ m$.$\Delta\lambda = |7.5 	imes 10^{-7} - 6.0 	imes 10^{-7}| = 1.5 	imes 10^{-7} \ m$.Given $\Delta	heta = 0.6 \ 	ext{radian}$.From $d = \frac{2\Delta\lambda}{\Delta	heta}$:$d = \frac{2 	imes 1.5 	imes 10^{-7}}{0.6} = \frac{3.0 	imes 10^{-7}}{0.6} = 5 	imes 10^{-7} \ m$.

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:

The angular spread of the central maximum in a diffraction pattern does not depend on . . . . . . .

In a single slit diffraction pattern:

$A$ light wave is incident normally on a slit of width $24 \times 10^{-5} \, cm$. The angular position of the second dark fringe from the central maximum is $30^\circ$. What is the wavelength of light in $\mathring{A}$?

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