Question

The diameter of the objective lens of microscope makes an angle \(\beta\) 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 \(n\sin 2\beta\)
• D. Increases with increasing value of \(\frac{1}{n\sin 2\beta}\)

Hint:

Resolving power of a microscope is \(\frac{2n\sin\theta}{1.22\lambda}\), which grows with n and with the half-angle of the cone.

Answer 1

The Correct Option is C

Resolving power of a microscope: \(RP = \frac{1}{d_{min}} = \frac{2n\sin\theta}{1.22\lambda}\), where \(\theta\) is the half-angle of the cone of light accepted by the objective.
So RP increases when the refractive index n of the medium increases and when \(\sin\theta\) increases, i.e. RP increases with increasing value of n times the sine of the cone angle.
Option (1) (decreasing n) and option (2) (decreasing angle) both reduce RP, and option (4) (increasing \(1/(n\sin 2\beta)\)) means decreasing \(n\sin2\beta\), which also reduces RP. Only option (3) states the correct trend.
Answer: (3) Increases with increasing value of \(n\sin 2\beta\)