The aperture of the objective of a telescope is $24.4 \, cm$. What is the resolving power of this telescope if light of wavelength $2440 \, \mathring{A}$ is used to view the object?

By increasing the aperture of the objective lens,how do the wavelength of light,the focal length of the objective lens,and the resolving power of an astronomical telescope change,respectively?

We use a simple microscope to magnify an object. The microscope has a numerical aperture of $\sin \alpha = 0.24$. The object is so small that the resolving power of the microscope is fully utilized. If the diameter of the eye's pupil is $d = 4.0 \ mm$ and the least distance of distinct vision is $D = 25 \ cm$,what is the minimum magnifying power of the microscope?

The wavelengths of light used in an optical instrument are ${\lambda _1 = 4000 \ \text{\AA}}$ and ${\lambda _2 = 5000 \ \text{\AA}}$. The ratio of their respective resolving powers (corresponding to ${\lambda _1}$ and ${\lambda _2}$) is:

Wavelengths of light used in an optical instrument are $\lambda_1 = 4000 \; \mathring{A}$ and $\lambda_2 = 5000 \; \mathring{A}$. The ratio of their respective resolving powers (corresponding to $\lambda_1$ and $\lambda_2$) is:

$A$ telescope has an objective lens of $10\; m$ diameter and is situated at a distance of $1\; km$ from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is $5000\; \text{\AA}$, is of the order of: