The diameter of a human eye lens is 2,mm. Wha | vedclass

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(C) The limit of resolution for an aperture is given by the formula $	heta = \frac{y}{D} = 1.22 \frac{\lambda}{d}$.For small angles,we consider the R

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$1.25\,cm$

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(C) The limit of resolution for an aperture is given by the formula $	heta = \frac{y}{D} = 1.22 \frac{\lambda}{d}$.For small angles,we consider the Rayleigh criterion $\frac{y}{D} \approx \frac{\lambda}{d}$.Given:Diameter of lens $d = 2\,mm = 2 	imes 10^{-3}\,m$.Distance $D = 50\,m$.Wavelength $\lambda = 5000\,\mathring{A} = 5 	imes 10^{-7}\,m$.Substituting the values:$y = \frac{\lambda D}{d} = \frac{5 	imes 10^{-7} 	imes 50}{2 	imes 10^{-3}}$.$y = \frac{250 	imes 10^{-7}}{2 	imes 10^{-3}} = 125 	imes 10^{-4}\,m$.$y = 1.25 	imes 10^{-2}\,m = 1.25\,cm$.

The diameter of a human eye lens is $2\,mm$. What will be the minimum distance between two points to resolve them,if they are situated at a distance of $50\,m$ from the eye? The wavelength of light used is $5000\,\mathring{A}$.

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