Question

The resolving power of a compound microscope:
A. Decreases on decreasing the diameter of objective and increases on increasing the focal length.
B. Decreases on increasing the diameter of objective and increases on increasing the focal length.
C. Increases the diameter of objective and does not depend on the focal length.
C. increases on increasing the diameter of objective and decreases on increasing the focal length.

Answer 1

Hint: Study the construction and working process of a compound microscope. Learn about how different parts work to give the final result. Obtain the formula for resolving power and try to define how the resolving power depends on other quantities.

Complete step by step answer:
The resolving power of a compound microscope can be defined as the smallest distance of the object, of which the microscope can form a separate image.
A compound microscope can be defined as a microscope which uses two sets of lenses to produce a higher magnification than a simple microscope. It is called a compound microscope because it uses a compound lens system to produce magnified images.
The compound microscope consists of two optical parts called the objective lens and the ocular lens.
Now resolving power of the compound microscope is given by the reciprocal of the minimum distance between two objects that are to be resolved by the microscope.
$d_{min}={\displaystyle \frac{1.22\lambda }{2n\sin \theta }}$
Where $\lambda$ is the wavelength of the incoming light, n is the refractive index of the medium separating objective and aperture.
Now, resolving power is given by,
$\begin{array}{rl}& R.P={\displaystyle \frac{1}{d_{min}}}={\displaystyle \frac{2n\sin \theta }{1.22\lambda }}\\ & R.P.\propto {\displaystyle \frac{n\sin \theta }{\lambda }}\end{array}$
So, the resolving power of the compound microscope is directly proportional to value of $n\text{ and sin}\theta$and inversely proportional to the value of $\lambda$.
Now, as the diameter of the objective lens increases the value of $\theta$ will increase and the value of $\sin \theta$ will also increase. Since, resolving power of the telescope is directly proportional to the value of $\sin \theta$, resolving power will increase with increasing diameter of the objective.
Again, the resolving power of the compound microscope is independent of the focal length.
So, the correct option is (C).

Note: The resolving power of the compound microscope depends on other quantities as, with the increase in wavelength of light the resolving power decreases and with the increase in refractive index the resolving power increases.