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  • For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for the light of 5000 \AA and electrons accelerated through 100V used as the illuminating substance.

For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for the light of  5000 \AA and electrons accelerated through 100V used as the illuminating substance.

 

Answers (1)

5000 \AA = 5000 \times 10^{-10} m\\ \\ \frac{1}{d} = \frac{2 \sin \beta}{1.22 \lambda } \\\\ d\;min = \frac{1.22 \lambda}{2 \sin \beta}\\d_{min}=\frac{1.22\times5000\times10^{-10}}{2sin\beta}

When we use 100V light,

\lambda_d =\frac{1.27}{\sqrt{V}}nm=\frac{1.27}{\sqrt{100}}nm

d' min =\frac{1.22\lambda_d }{2 \sin \beta}

d'min = \frac{1.22 \times 1.27 \times 10 ^{-10} }{2 \sin \beta}

The required ratio =\frac{d \;min}{d'min} = \frac{1.22 }{5000} =0.244 \times 10^{-3}

Posted by

Gurleen Kaur

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