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Question

(A) The resolving power of the microscope determines the smallest detail it can resolve. The minimum resolvable distance $d_{\min}$ is given by $d_{\m

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) The resolving power of the microscope determines the smallest detail it can resolve. The minimum resolvable distance $d_{\min}$ is given by $d_{\min} = \frac{0.61 \lambda}{\sin \alpha}$.The angular resolution of the microscope is $	heta_0 = \frac{d_{\min}}{D} = \frac{0.61 \lambda}{D \sin \alpha}$,where $D = 25 \ cm = 250 \ mm$ is the least distance of distinct vision.The human eye has an angular resolution limit $	heta = \frac{1.22 \lambda}{d}$,where $d = 4.0 \ mm$ is the diameter of the pupil.To see the smallest detail resolved by the microscope,the magnified angle $	heta$ must match the eye's resolution limit. The magnifying power $m$ is the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the near point.$m = \frac{	heta}{	heta_0} = \frac{1.22 \lambda / d}{0.61 \lambda / (D \sin \alpha)} = \frac{2 D \sin \alpha}{d}$.Substituting the values: $m = \frac{2 	imes 250 \ mm 	imes 0.24}{4.0 \ mm} = \frac{120}{4} = 30$.

Similar Questions

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