Physics Topics are often described using mathematical equations, making them precise and quantifiable.
What is Refractive Index? What is the Derivation of Snell’s Law Formula?
Introduction
Definition: When a ray of light travelling in one medium enters another medium obliquely, the ray changes its direc-tion at the interface. This phenomenon is known as refrac-tion of light.
Let CD be the plane of separation of the media—air and glass [Fig.]. It is called refracting surface. The ray AO incident obliquely at O, changes direction after refraction and goes along the line OB. NON’ is drawn perpendicular to the surface of separation. The perpendicular to CD, NON’ is called normal. AO is the incident ray and OB is the refracted ray. Angle between the incident ray and the normal to the surface of sepa-ration at the point of incidence is called the angle of incidence
(i) Angle between the refracted ray and the normal to the sur-face of separation is called the angle of refraction (r).
Refraction from rarer to denser medium: if the ray travels from an optically rarer medium to a denser medium, say from air to glass, the refracted ray bends towards the normal [Fig.]. Here i > r.
Refraction from denser to rarer medium: If the ray travels from denser to rarer medium, say from glass to air, the refracted ray bends away from the normal [Fig.], Here r > i.
Laws Of Refraction
Refraction of light is governed by the following two laws, called laws of refraction.
i) The incident ray, the refracted ray and the normal at the point of incidence lie on the same plane.
ii) The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant. This constant depends on the nature of the two concerned media and on the colour of ray used.
This second law of refraction is known as Snell’s law, named after Willebrord Snellius (1580-1626).
Refractive Index
If i is the angle of incidence and r is the angle of refraction, then according to the second law,
\(\frac{\sin i}{\sin r}\) = µ (pronounced as ‘mu’) = constant,
This constant is called the refractive index of the second medium with respect to the first medium.
The value of the refractive index depends on
- nature of the two concerned media and
- colour of the incident light.
Whatever may be the value of the angle of incidence, the value of the refractive index will remain constant if the colour of the incident light (i.e., frequency) and the two media remain unchanged.
Obviously µ has no unit.
Normal incidence: If a ray of light is incident perpendic-ularly on a refracting surface, then i = 0. According to Snell’s law,
µsinr = sin 0 = 0 or, r = 0
So, the ray suffers no deviation.
Relative refractive index: when light passes from a medium a into another medium b, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is called the refractive index of medium b with respect to the medium a . It is denoted by aµb i.e.,
aµb = \(\frac{\sin i}{\sin r}\)
[i = angle of incidence, r = angle of refraction]
This refractive index is called the relative refractive index.
According to the principle of reversibility of light—a ray of light will follow exactly the same path if its direction of travel is reversed.
Following this principle we can say that the ray BO in medium b when incidents at an angle r on the interface of the second medium a, refracts at an angle i along the path OA [Fig.]. Compare this figure with Fig., it is the just opposite phenom-enon.
Thus, bµa = \(\frac{\sin r}{\sin i}\)
Here bµa is the refractive index of medium a with respect to medium b.
So, aµb = \(\frac{\sin i}{\sin r}\) × \(\frac{\sin r}{\sin i}\) = 1
or, aµb = \(\frac{1}{b^{\mu_a}}\)
For example, if the refractive index of water with respect to air is \(\frac{4}{3}\), then refractive index of air with respect to water is \(\frac{3}{4}\).
Absolute refractive index: When light refracted from vacuum to another medium, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is called the abso-lute refractive index of the medium.
If the angle of incidence is i and the angle of refraction is r, then absolute refractive index of the medium,
µ = \(\frac{\sin i}{\sin r}\)
Therefore, ralative refractive index of a medium with respect to vacuum is the absolute refractive index of that medium. Obvi-ously, refractive index of vacuum is 1. In general, refractive index of a medium relative to air medium is considered as the refractive index of that medium. But it is not actual absolute refractive index of the medium.
It is an experimental fact that the difference in the values of the refractive index of a medium with respect to air and its absolute refractive index is very small. For example, at STP, absolute refractive index of air is 1.0002918 . So the refractive index of any medium with respect to air may be considered as its abso-lute refractive index.
e.g., refractive index of glass is 1.5—it means that refractive index of glass with respect to air is 1.5. At STP, absolute refractive index of air is 1.0002918 and refractive index of glass with respect to air is 1.5. Thus, absolute refractive index of glass = \(\frac{1.5}{1.0002918}\) = 1.49956 ≈ 1.5 . Absolute refractive index of a medium is denoted by µ. If there is more than one medium then µ1, µ2, µ3 are used.
Optical Density of a Medium
If the absolute refractive index (µ1) of any medium is greater than that of another medium (µ2), then the first medium is called optically denser and the second medium is called optically rarer. So, if µ1 > µ2 then medium 1 is optically denser with respect to medium 2 i.e., medium 2, is optically rarer with respect to medium 1. Optical density of a medium has no relation with its physical density or specific gravity. For example, specific gravity of turpentine oil is 0.87 and that of water is 1. But refractive index of turpentine oil is 1.47 and that of water is 1.33.
The refractive index of a medium depends on the colour of the incident light. It is greater for blue or violet than for red. The refracted ray in case of violet light bends more than in case of red light. The refractive index for yellow light is midway between these two.
So unless otherwise stated, refractive index of a medium refers to yellow light.
Refractive Indices of a Few Substances
Refractive Index and Related Terms
Relation between velocity of light and refractive index: According to the wave theory of light, velocity of light is different in different media. If µ is the absolute refractive index of a medium, then
µ \(=\frac{\text { velocity of light in vacuum }(c)}{\text { velocity of light in that medium }(\nu)}\)
∴ Speed of light in free space = µ × speed of light in that medium
For any medium µ > 1, so speed of light in vacuum is greater than that in any medium. Thus, speed of light is maxium in vac-uum.
Accordingly, if aµb is the refractive index of the medium b with respect to the medium a, then
aµb \(=\frac{\text { velocity of light in medium } a}{\text { velocity of light in medium } b}\)
= \(\frac{v_a}{v_b}\)
If the medium b is.denser than medium a, then aµb > 1 and in this case va > vb.
Velocity of light in a denser medium is less than the velocity of light in a rarer medium. Velocity of light in a medium decreases with the increase of its refractive index.
i) Thus if velocity of light in medium b is lesser, i.e., medium b is optically denser then, aµb > 1 i.e., sin i > sin r or i > r. Thus, angle of refraction is less than angle of incidence, so the refracted ray bends towards the normal.
ii) If velocity of light in medium b is more, i.e., medium b is optically rarer then, aµb < 1 i.e., sin i < sin r or i < r. Thus, angle of refraction is greater than angle of incidence so, the refracted ray bends away from the normal.
From above discussion it is clear that refraction occurs due to variation in speed of light in different media.
Relation between relative refractive index and absolute refractive index: If µa and µb are absolute refractive indices of media a and b respectively.
µa = \(\frac{c}{v_a}\) and µb = \(\frac{c}{v_b}\)
So relative refractive index of medium b with respect to medium a,
aµb = \(\frac{v_a}{v_b}\) = \(\frac{\frac{c}{v_b}}{\frac{c}{v_a}}\) = \(\frac{\mu_b}{\mu_a}\)
Relation of wavelength of light with refractive index: The relative refractive index between two media depends on the wavelength of light. Cauchy’s equation for the dependence of refractive index on wavelength of light is
µ = A + \(\frac{B}{\lambda^2}\)
Here A and B are two constants; their values are different in dif-ferent media. It is evident that refractive index of a medium decreases if wavelength of light increases. In Fig., the graph shows the variation of µ with λ of BK7 glass.
In a medium apart from free space different coloured light travels in different speed. In a particular medium red travels the fastest and violet travels the slowest. Thus refractive index of any medium for red colour is the lowest and that for violet colour is the highest. This is the reason why white light is dispersed.
Relation of temperature with refractive index: Generally, the refractive index of a medium decreases if the temperature of the medium increases. For a solid medium this change is small, for a liquid it is moderate and for a gas it is remarkable.
It is to be remembered that, velocity, intensity and wave-length of light change due to refraction but its frequency and phase remain unchanged.
Generalised Form of Snell’s Law
Let AB be the surface of separation of two media 1 and 2 [Fig.]. Medium 2 is denser
and medium 1 is rarer. PO is the incident ray at the point O on the surface of separation and OQ is the refracted ray at the point O.
Let angle of incidence = i1, angle of refraction = i2.
By Snell’s law, = \(\frac{\sin i_1}{\sin i_2}\) = i1µ2; we know, 1µ2 = \(\frac{\mu_2}{\mu_1}\)
∴ \(\frac{\sin i_1}{\sin i_2}\) = \(\frac{\mu_2}{\mu_1}\) or,
µ1 sin i1 = µ2 sin i2 ….. (1)
So for n number of media It can be written as
µ1 sin i1 = µ2 sin i2 = ……. = µn sin in ….. (2)
This equation is known as the generalised form of Snell’s law.
Numerical Examples
Example 1.
A ray of light is Incident from water on the surface of separation of air and water at an angle 30°. Calculate the angle of refraction in air. µ of water = \(\frac{4}{3}\)
Solution:
Let the angle of refraction of the ray of light in air be r [Fig.]. Since the ray of light is refracted from water to air,
Example 2.
A ray of light is Incident on a block of glass in such a way that the angle between the reflected ray and the refracted ray is 90°. Determine the relation between the angle of incidence and the refractive index of glass.
Solution:
Here angle of incidence = i
∴ angle of reflection = i,
angle of refraction = r
angle between the reflected ray and the refracted ray = 90°
According to the Fig.,
i + 90° + r = 180°
or, r = 90° – i
Refractive index of glass,
µ = \(\frac{\sin i}{\sin r}\) = \(\frac{\sin i}{\sin \left(90^{\circ}-i\right)}\) = \(\frac{\sin i}{\cos i}\) = tan i
Example 3.
Refractive Index of glass is 1.5 and refractive index of water is 1.33. If the velocity of light in glass is 2 × 108 m ᐧ s-1, what is the velocity of light in water?
Solution:
Example 4.
A monochromatic ray of light is refracted from vacuum to a medium of refractive index µ. Determine the relation of the wavelengths of light in vacuum and in glass.
Solution:
µ = \(\frac{c}{v}\) = \(\frac{n \lambda_0}{n \lambda}\) = \(\frac{\lambda_0}{\lambda}\)
So, λ0 = µλ
[Here, c and v are the velocities of light in vacuum and in the medium respectively; n = frequency of light, which remains unchanged on refraction; λo, λ = wavelengths in vacuum and in the medium respectively.]
Example 5.
If a ray of light is Incident on a plate inside water at an angle of 45°, what is the angle of refraction inside the plate? Given that the absolute refractive indices of the plate and water are 1.88 and 1.33 respectively.
Solution:
Let the angle of refraction inside the plate be r.
Here, \(w^{\mu_g}\) = \(\frac{\sin i}{\sin r}\) or, \(\frac{\mu_g}{\mu_w}\) = \(\frac{\sin i}{\sin r}\)
or, \(\frac{1.88}{1.33}\) = \(\frac{\sin 45^{\circ}}{\sin r}\) [given µg = 1.88, µw = 1.33]
or, sin r = \(\frac{1}{\sqrt{2}} \times \frac{1.33}{1.88}\) = 0.5 = \(\frac{1}{2}\) = sin 30°
∴ r = 30°
Example 6.
How much time will sunlight take to pass through the glass window of thickness 4 mm? µ of glass = 1.5.
Solution:
Velocity of sunlight in vacuum or in air,
c = 3 × 108 m ᐧ s-1
Thus velocity in a medium of refractive index µ,
v = \(\frac{c}{\mu}\)
So to cross a thickness d, the time taken by light,
t = \(\frac{d}{v}\) = \(\frac{d \mu}{c}\) = \(\frac{\left(4 \times 10^{-3}\right) \times 1.5}{3 \times 10^8}\)
[here d = 4 mm = 4 × 10-3 m]
= 2 × 10-11 s
Example 7.
Green Light of wavelength 5460 A is incident on an air-glass interface. If the refractive Index of glass is 1.5 what will be the wavelength of light in glass?
Solution:
Wavelength of the light in air, λ0 = 5460 A
Refractive index of glass with respect to air, µ = 1.5
If the wavelength of the light in glass is λ, then
µ = \(\frac{\lambda_0}{\lambda}\)
or, λ = \(\frac{\lambda_0}{\mu}\) = \(\frac{5460}{1.5}\) = 3640 A