Question

Calculate the resolving power of a microscope with cone angle of light falling on the objective equal to ${60}^{\circ }.$ Take $\lambda =600\text{nm}$ and the refractive index of transparent medium between the object and the objective lens of the microscope, $\mu =4/3.$

• A. $1.82\times {10}^6{\text{m}}^{-1}$
• B. $2\times {10}^6{\text{m}}^{-1}$
• C. $3.22\times {10}^6{\text{m}}^{-1}$
• D. $1.22\times {10}^6{\text{m}}^{-1}$

Answer 1

The correct option is A $1.82\times {10}^6{\text{m}}^{-1}$

Given :

$\lambda =600\text{nm};\mu =\frac{4}{3}2\theta ={60}^{\circ }\Rightarrow \theta ={30}^{\circ }$


The resolving power of the microscope is given by,

$RP=\frac{2\mu \sin \theta }{1.22\lambda }$

$\Rightarrow RP=\frac{2\times \left(\frac{4}{3}\right)\times \sin {30}^{\circ }}{1.22\times 600\times {10}^{-9}}$

$\therefore RP=1.82\times {10}^6{\text{m}}^{-1}$

Hence, $\left(\text{A}\right)$ is the correct answer.