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 and electrons accelerated through 100V used as the illuminating substance.
✓
Answer
Resolving power of a microscope $=\frac{2\sin\beta}{1.22\lambda}$ where $\mu$ is refractive index of the medium, $\lambda$ is the wavelength of light and $\beta$ is the angle subtended by the objective at the object.
Now, $\frac{1}{\text{d}}=\frac{2\sin\beta}{1.22\lambda}$
$\Rightarrow\ \text{d}_\text{min}=\frac{1.22\lambda}{2\sin\beta}$
For the light of wavelength 5000A,
$\text{d}_\text{min}=\frac{1.22\times5000\times10^{-10}}{2\sin\beta}\Big[\because\ 1\text{A}=10^{-10}\text{m}\Big]\ .....(\text{i})$
Now, the de-Broglie wavelenght is given by $\lambda=\frac{12.27}{\sqrt{\text{V}}}$
Substituting V = 100 in $\lambda=\frac{12.27}{\sqrt{\text{V}}}$, we get
$\frac{12.27}{\sqrt{100}}=1.227\times10^{-10}\text{m}$
$\therefore\ \text{d}'_\text{min}=\frac{1.22\times1.227\times10^{-10}}{2\sin\beta}$
Ratio of the least separation
$\frac{\text{d}_\text{min}}{\text{d}'_\text{min}}=\frac{\frac{1.22\times5000\times10^{-10}}{2\sin\beta}}{\frac{1.22\times1.227\times10^{-10}}{2\sin\beta}}$
$=\frac{5000}{1.227}$
$=4075.$
Now, $\frac{1}{\text{d}}=\frac{2\sin\beta}{1.22\lambda}$
$\Rightarrow\ \text{d}_\text{min}=\frac{1.22\lambda}{2\sin\beta}$
For the light of wavelength 5000A,
$\text{d}_\text{min}=\frac{1.22\times5000\times10^{-10}}{2\sin\beta}\Big[\because\ 1\text{A}=10^{-10}\text{m}\Big]\ .....(\text{i})$
Now, the de-Broglie wavelenght is given by $\lambda=\frac{12.27}{\sqrt{\text{V}}}$
Substituting V = 100 in $\lambda=\frac{12.27}{\sqrt{\text{V}}}$, we get
$\frac{12.27}{\sqrt{100}}=1.227\times10^{-10}\text{m}$
$\therefore\ \text{d}'_\text{min}=\frac{1.22\times1.227\times10^{-10}}{2\sin\beta}$
Ratio of the least separation
$\frac{\text{d}_\text{min}}{\text{d}'_\text{min}}=\frac{\frac{1.22\times5000\times10^{-10}}{2\sin\beta}}{\frac{1.22\times1.227\times10^{-10}}{2\sin\beta}}$
$=\frac{5000}{1.227}$
$=4075.$
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