Ncert Exercises & Solutions

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 $\overset{o}{A}$ and electrons accelerated through 100V used as the illuminating substance.

Hint: The resolution power of the microscope depends on the wavelength of the light.

Step 1: Find the least distance for the microscope.

We know that,

Resolving power $= \frac{1}{d} = \frac{2 sinβ}{1.22 λ} \Rightarrow d_{\min} = \frac{1.22 λ}{2 \sin β}$

where $λ$ is the wavelength of light and $β$ is the angle subtended by the objective at the object.

For the light of wavelength 5000 $\overset{o}{A}$ ,

$d_{\min} = \frac{1.22 \times 5000 \times 10^{- 10}}{2 \sin β}$

Step 1: Find the least distance for the electrons.

For electrons accelerated through 100 V, the de-Broglie wavelength;

$λ = \frac{12.27}{\sqrt{V}} = \frac{12.27}{\sqrt{100}} = 0.12 \times 10^{- 9} m$
$d_{\min}' = \frac{1.22 \times 0.12 \times 10^{- 9}}{2 \sin β}$

The ratio of the least separation;

$\frac{d_{\min}'}{d_{\min}} = \frac{0.12 \times 10^{- 9}}{5000 \times 10^{- 10}} = 0.24 \times 10^{- 3}$

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