Question

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

Answer 1

Resolving power:

It is defined as the inverse of the distance or angular separation between two objects which can be just resolved when viewed through the optical instrument.

Given, ${\lambda }_1=5000A^{\circ },\text{Voltage}=100V$

For the light of wavelength $5000A^{\circ }$, Resolving power.

$R=\frac{1}{d}=\frac{2\sin \alpha }{1.22\lambda }\Rightarrow d_{min}=\frac{1.22\lambda }{2\sin \alpha }$

$\lambda =\text{wavelength of light}$.
$\alpha =\text{Angle subtended by the object}$
$d_{min}=\frac{1.22\times 5000\times {10}^{-10}}{2\sin \alpha }$
For electrons acceralted through $100V$, the de-Brogile wavelength,
$\lambda =\frac{12.27}{\sqrt{V}}=\frac{12.27}{\sqrt{100}}=0.12\times {10}^{-9}m$

For the light of wavelength $0.12\times {10}^{-9}m$, Resolving power,
$d_{min}^{{}^{\prime }}=\frac{1.22\times 0.12\times {10}^{-9}}{2\sin \alpha }$
Ratio of the least separtion,
$\frac{d_{min}^{{}^{\prime }}}{d_{min}}=\frac{0.12\times {10}^{-9}}{5000\times {10}^{-10}}=0.2\times {10}^{-3}$

Final Answer: $2\times {10}^{-4}$