Question

Find the dimensions of the refractive index.

Answer 1

Hint: Try to understand the definition of refraction and then of refractive index. Then try to break it in its components and then you can find the dimension.

Complete step by step answer:
When a beam of light encounters the interface of two mediums, a part of the light is reflected back to the first medium and the other passes through the second medium. The direction of propagation of an oblique incident (at an angle less than 90 degree) light beam changes at the interface of the two media. This is known as reflection of light.
Snell’s law describes the changes as the laws of refraction:
The incident ray, the refracted ray and the normal to the interface at the point of refraction all lie in the same plane.
The ratio of sine of incident angle to the sign of refracted angle is a constant for the given media.
$\dfrac{\sin i}{\sin r}={{n}_{21}}$
Where, i is the angle of incidence and r is the angle of refraction.
${{n}_{21}}$ is a constant and is called the refractive index of the second medium with respect to the first medium.
Refractive index can be defined as the ratio of velocity of light in the two media through the interface of which light is passed.
${{n}_{12}}=\dfrac{1}{{{n}_{21}}}$
Where, ${{n}_{12}}$ is the refractive index of the first medium with respect to the second medium.
So, we can define the refractive index as the ratio of velocities in the two media. So refractive index is a unitless quantity (ratio or just a number).
Hence, we can say that the refractive index is dimensionless.
Dimension of refractive index is $\left[ {{M}^{0}}{{L}^{0}}{{T}^{0}} \right]$

Note: Relative refractive index can be defined as the ratio of velocity of light in the given media with respect to the velocity of light in vacuum. We can write $n=\dfrac{c}{v}$, where n is the relative refractive index of the medium, c is the speed of light in vacuum and v is the speed of light in the given medium.