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(A) The angular resolution of the eye is given by $	heta = 1.22 \frac{\lambda}{D}$,where $\lambda$ is the wavelength and $D$ is the diameter of the p

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$25$

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(A) The angular resolution of the eye is given by $	heta = 1.22 \frac{\lambda}{D}$,where $\lambda$ is the wavelength and $D$ is the diameter of the pupil.For an object at distance $x$ and size $h$,the angle subtended is $	heta = \frac{h}{x}$.Since the object is 'just resolved',we have $\frac{h}{x} \propto \frac{\lambda}{D}$.Thus,$x \propto \frac{hD}{\lambda}$. Assuming the object size $h$ remains constant,we have $\frac{x_2}{x_1} = \frac{D_2}{D_1} 	imes \frac{\lambda_1}{\lambda_2}$.Given $x_1 = 25 \ cm$,$\lambda_1 = 500 \ nm$,$D_1 = 1 \ mm$,$\lambda_2 = 400 \ nm$,and $D_2 = 0.8 \ mm$.Substituting the values: $x_2 = 25 	imes \frac{0.8}{1} 	imes \frac{500}{400} = 25 	imes 0.8 	imes 1.25 = 25 	imes 1 = 25 \ cm$.

If an object kept at the least distance of distinct vision is just resolved with light of wavelength $500 \ nm$ and a pupil of diameter $1 \ mm$,at what distance will the object be just resolved if the wavelength is $400 \ nm$ and the pupil diameter is $0.8 \ mm$ (in $cm$)?

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