When the wavelengths of light used in optical instruments $A$ and $B$ are $4500 \, \text{Å}$ and $6000 \, \text{Å}$ respectively, the ratio of the resolving power of $A$ to $B$ will be:

Assuming the human pupil to have a radius of $0.25 \ cm$ and a comfortable viewing distance of $25 \ cm$,the minimum separation between two objects that the human eye can resolve at $500 \ nm$ wavelength is..... $\mu m$.

The human eye has an approximate angular resolution of $\phi = 5.8 \times 10^{-4} \, rad$ and a typical photoprinter prints a minimum of $300 \, dpi$ (dots per inch, $1 \, inch = 2.54 \, cm$). At what minimal distance $z$ should a printed page be held so that one does not see the individual dots?

The diameter of the objective lens of a 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

$A$ microscope was initially placed in air (refractive index $1$). It is then immersed in oil (refractive index $2$). For a light whose wavelength in air is $\lambda$,calculate the change of the microscope's resolving power due to oil and choose the correct option.

Which of the following is the smallest size grain that can be seen using a normal microscope using visible light $(400 \ nm < \lambda < 700 \ nm)$ (in $nm$)?