Question

Resolving power of a microscope depend upon,
(A) The focal length and aperture of the eye lens
(B) The focal lengths of the objective and the eye lens
(C) The apertures of the objective and the eye lens
(D) The wavelength of light illuminating the object

Answer 1

Hint: Resolving power is the power of an optical device to distinguish between two firmly placed objects which are located at a distance, and produce their images. The resolving power is inversely proportional to the distance between two objects.

Formula Used: The following formulas are used to solve this question.
 $ \Rightarrow {d_{\min }} = \dfrac{{1.22\lambda }}{{2\mu \sin \theta }} $ where $ {d_{\min }} $ is the minimum separation between two objects, $ \lambda $ is the wavelength of light illuminating the object, $ \mu $ is the refractive index of the medium and $ \theta $ is half the angle of the lens, the value of which depends on focal length and diameter of the lens.
 $ \Rightarrow R.P. = \dfrac{1}{{{d_{\min }}}} = \dfrac{{2\mu \sin \theta }}{{1.22\lambda }} $ where $ R.P. $ is the resolving power of the microscope.

Complete step by step answer
Resolving Factor of a microscope is defined as the inverse of the distance or angular separation between two objects which can be just resolved when viewed through the optical instrument.
More is the resolving power of a device, better will be the image quality and clarity, and things can be seen clearly.
Resolving power is different from magnification. While magnification only magnifies images, i.e. it enlarges images, resolving power enables us to see two different closely spaced objects located at a distance. Magnification and resolution are inversely proportional to each other. So, when magnification increases, the resolution decreases, and vice versa.
In case of microscopes, resolving power is inversely proportional to the distance between two objects.
The minimum separation between two objects that are to be resolved by a microscope is given by,
 $ \Rightarrow {d_{\min }} = \dfrac{{1.22\lambda }}{{2\mu \sin \theta }} $, where $ {d_{\min }} $ is the minimum separation between two objects, $ \lambda $ is the wavelength of light illuminating the object, $ \mu $ is the refractive index of the medium and $ \theta $ is half the angle of the lens, the value of which depends on focal length and diameter of the lens.
The resolving power is therefore, given by,
 $ \Rightarrow R.P. = \dfrac{1}{{{d_{\min }}}} = \dfrac{{2\mu \sin \theta }}{{1.22\lambda }} $, where $ R.P. $ is the resolving power of the microscope.
 $ \therefore R.P.\alpha \dfrac{{\mu \sin \theta }}{\lambda } $
Hence, among the options given, resolving power depends on the wavelength of light illuminating the object.

The correct answer is Option D.

Note
Resolving power doesn't have any SI unit. This is because the resolving power is the ratio of a mean wavelength of a pair of spectral lines and the difference of wavelength between them. Since both the quantities have the same unit, the resolving power has no unit. The wavelength of light, refractive index, and angular aperture are the significant factors that affect the resolving power.