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**Hint**in this question, we need to find the relation between the resolving power of the microscope and the refractive index of the glass used. It is given that the objective lens makes an angle $\beta $ with the focus of the microscope. This means that the total aperture angle would be equal to 2 $\beta $.

**Complete step by step solution**

We know that the minimum separation between to objects for them to be successfully resolved by a microscope is given by the following expression:

${d_{\min }} = \dfrac{{1.22\lambda }}{{2n\sin \theta }}$

Where, n is the refractive index of the material and \[\lambda \] is the wavelength of the light used.

The resolving power is inversely proportional to this minimum separation between the two objects.

\[ \Rightarrow R.P. = \dfrac{1}{{{d_{\min }}}}\]

Now, we will substitute the value of \[{d_{min}}\] into this expression for resolving power,

\[

\Rightarrow R.P. = \dfrac{1}{{\dfrac{{1.22\lambda }}{{2n\sin \theta }}}} \\

\Rightarrow R.P. = \dfrac{{2n\sin \theta }}{{1.22\lambda }} \\

\]

It is given to us that the diameter of the objective lens of the microscope makes an angle ‘ $\beta $ ’ at the focus of the microscope.

Hence, $\theta = 2\beta $ .

On substituting this, the final expression for resolving power of microscope becomes,

\[ \Rightarrow R.P. = \dfrac{{2n\sin 2\beta }}{{1.22\lambda }}\]

From this expression we can deduce that our resolving power of the microscope is directly proportional to \[n\sin 2\beta \]. So, the resolving power increases with increase in \[n\sin 2\beta \] .

**Therefore, option (C) is correct.**

**Note**The physical significance that one must know here, is that the resolving power of a microscope is the ability of the microscope to form two separate images for two very closely placed objects.

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