This article deals with the refractive index formula and its derivation. Refractive index refers to a value which has common usage in optical science. Furthermore, it has no unit. Refractive Index indicates a material’s ability to refract light. In order to properly understand the concept of refractive index, one must become familiar with the concept of refraction. Simply speaking, refraction takes place when light or any other radiation changes medium.

**What is Refractive Index?**

Refractive index refers to the ratio between the speed of an electromagnetic wave in a vacuum and its speed in another medium. It indicates the degree of refraction that would occur in a particular material. Furthermore, the refractive index through a vacuum is 1. One can define the normal as the perpendicular line to the surface of the interface. In such a case, a higher value of refractive index would certainly mean that the bend of the ray is towards the normal.

Refractive index refers to the measure of the bending of a ray of light when it passes from one medium to another medium. Most refractive index values which make use of visible light are between 1 and 2. Furthermore, refractive index values may be significantly higher in the case of infrared waves.

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**Refractive Index Formula**

The refractive index formula is given as follows:

n = \(\frac{c}{v}\)

where,

n = refractive index of the medium

c = speed of light when in vacuum

v = speed of light when in medium

**Refractive Index Formula Derivation**

First off all, one must apply the Snell’s law as follows:

For a ray which goes from the air to a medium

Nasin i = nmsin r

Here nm refers to the refractive index of the medium

Na refers to the refractive index of air

i refers to the angle of incidence

r refers to the angle of refraction

Now one must nm the subject of the above equation

Nm = \(\frac{n_{a}sin i}{sin r}\)

Also, na = 1

Nm = \(\frac{sin i}{sin r}\)

Above all, nm = \(\frac{c}{v}\) = \(\frac{speed of light in the vacuum}{speed of light in the medium}\)

**Solved Questions on Resistivity Formula**

Q1. A ray of light is transient through a medium to another medium. The angle of the incident comes out to be 30 degrees. Also, the angle of refraction happens to be 50 degrees. Find out the refractive index of the second medium?

Answer: The parameters which are known are:

Angle of incidence (‘i’) = 30 degrees

Angle of refraction (‘r’) = 50 degrees

The refractive index formula is as follows

n = \(\frac{sin i}{sin r}\)

n = \(\frac{sin 30}{sin 50}\)

Finally, n = \(\frac{0.5}{0.766}\) = 0.6527

Q2. Find out the refractive index of the medium such that the speed of light in the medium is 2 × 108 m/s?

Answer: The information we have is,

Speed of light in the medium(‘v’) = 2×108m/s

Speed of light in the vacuum(‘c’) = 3×108m/s

Now, one must apply the formula of refractive index,

So, n = \(\frac{c}{v}\)

n = \(\frac{3\times 108}{2\times 108}\)

n = 1.5

Typo Error>

Speed of Light, C = 299,792,458 m/s in vacuum

So U s/b C = 3 x 10^8 m/s

Not that C = 3 x 108 m/s

to imply C = 324 m/s

A bullet is faster than 324m/s

I have realy intrested to to this topic

m=f/a correct this

Interesting studies

It is already correct f= ma by second newton formula…