Question

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

• A. 0.0002
• B. 0.003575
• C. 0.004075
• D. 5075

Answer 1

The correct option is A 0.0002

Given,

The voltage by which electrons are accelerated is $100V$

The wavelength of given light is ${\lambda }_1=5000A^{\circ }$

Now the limit of resolution of the microscope is given as

$2d\sin \theta =\lambda$

Hence,

$2d_1\sin \theta ={\lambda }_1$

$2\sin \theta =\frac{{\lambda }_1}{d_1}$ ……. (1)

Now according to de Broglie wavelength equation

$\lambda =\frac{12.27}{\sqrt{V}}{A}^{\circ }$

Hence,

${\lambda }_2=\frac{12.27}{\sqrt{100}}=1.227A^{\circ }$

Also

$2\sin \theta =\frac{{\lambda }_2}{d_2}$ …… (2)

Comparing equation (1) and (2)

$\frac{{\lambda }_2}{d_2}=\frac{{\lambda }_1}{d_1}\Rightarrow \frac{d_2}{d_1}=\frac{{\lambda }_2}{{\lambda }_1}$

$\frac{d_2}{d_1}=\frac{1.227}{5000}\Rightarrow \frac{d_2}{d_1}=0.00024$

Therefore the ratio of least separation between two points to be distinguished by the microscope is $0.0002$ .