In a Fresnel diffraction arrangement,the scre | vedclass

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(C) The radius of the $n^{th}$ Fresnel zone is given by the formula $r_n = \sqrt{n b \lambda}$,where $b$ is the distance from the aperture to the scre

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$5/4$

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(C) The radius of the $n^{th}$ Fresnel zone is given by the formula $r_n = \sqrt{n b \lambda}$,where $b$ is the distance from the aperture to the screen and $\lambda$ is the wavelength of light.Given that the radius of the $4^{th}$ zone for wavelength ${\lambda _1}$ is equal to the radius of the $5^{th}$ zone for wavelength ${\lambda _2}$,we have:$r_4 = \sqrt{4 b \lambda_1}$$r_5 = \sqrt{5 b \lambda_2}$Since $r_4 = r_5$,we equate the expressions:$\sqrt{4 b \lambda_1} = \sqrt{5 b \lambda_2}$Squaring both sides,we get:$4 b \lambda_1 = 5 b \lambda_2$Dividing both sides by $b$ (where $b 
eq 0$):$4 \lambda_1 = 5 \lambda_2$Therefore,the ratio is:$\frac{\lambda_1}{\lambda_2} = \frac{5}{4}$

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