Question

The diameter of the objective lens of microscope makes an angle ${}^{\prime }\beta {}^{\prime }$ at the focus of the microscope. Further, the medium between the object and the lens is an oil of refractive index 'n'. Then the resolving power of the microscope.

• A. Increases with decreasing value of $n$
• B. Increases with decreasing value of $\beta$
• C. Increases with increasing value of $nsin2\beta$
• D. Increases with increasing value of $\frac{1}{nsin2\beta }$

Answer 1

Answer: (C)

Increases with increasing value of $nsin2\beta$

Resolving Power of the Microscope is given by $\frac{1}{d}=\frac{2nsin\theta }{\lambda }$

where $n$ is refractive index , $\theta$ is the angle of the cone and d is the distance between the object .

Here , The diameter of the objective lens of microscope makes an angle $\beta$ at the focus of the microscope.

Hence , $\beta =\theta /2$

Hence , resolving power is $\frac{2nsin2\beta }{\lambda }$