Question 161: Ray Optics

Question

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 microscope's resolving power due to oil and choose the correct option.

Accepted Answer

The resolving power of a microscope is determined by the Rayleigh criterion, which states that it is inversely proportional to the wavelength of light used in the medium in which the microscopy is being performed. Mathematically, the resolving power (RP) can be represented as:
$ RP = \frac{1}{\lambda_n} $
where $\lambda_n$ is the wavelength of light in the medium, which can be found using the formula:
$ \lambda_n = \frac{\lambda}{n} $
Here, $\lambda$ is the wavelength of light in vacuum (or air, for practical purposes, since their refractive indices are close enough), and $n$ is the refractive index of the medium.
In air, the refractive index $n = 1$, so the wavelength of light in air ($\lambda_{air}$) is equal to $\lambda$.
In oil, the refractive index $n = 2$, so the wavelength of light in oil ($\lambda_{oil}$) is $\lambda / 2$.
Therefore, the change in resolving power when moving from air to oil can be calculated as the ratio of resolving powers in oil to air:
$ \frac{RP_{oil}}{RP_{air}} = \frac{\lambda_{air}}{\lambda_{oil}} = \frac{\lambda}{\lambda / 2} = 2 $
This means that the resolving power in oil is twice that in air. Thus, the correct option is:
Option B: Resolving power will be twice in the oil than it was in the air.

Answer

Resolving power will be $$\frac{1}{4}$$ in the oil than it was in the air.

Answer

Resolving power will be twice in the oil than it was in the air.

Answer

Resolving power will be four times in the oil than it was in the air.

Answer

Resolving power will be $$\frac{1}{2}$$ in the oil than it was in the air.

Ray Optics

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 microscope's resolving power due to oil and choose the correct option.

Choose the correct answer: