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Fermat's principle is a part of ray optics that deals with the travelling of light between two points in such a way that the number of waves, which is the optical length between the points, is equal to that of the neighbouring path. This principle is an essential part of ray optics. This principle is used to study optical devices. It was developed to find the path travelled by light in the shortest time, that’s why Fermat's principle is also known as the Principle of Least Time. Meaning, that a ray of light passes from one point to another select a certain path along which the time taken is minimum. However, the time taken by the light ray is either maximum or minimum for the spherical surfaces.
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Key Terms: Fermat's Principle, Snell’s Law, Reflection, Wave Optics, Ray Optics, Refraction, Homogenous Medium, Prism
What is Fermat's Principle?
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Fermat's principle is the principle of least time, which is one of the principles which connect ray optics and wave optics. Therefore, Fermat formulated that light travelling through a path either takes the maximum or minimum time. He considered the quickest time to cover the path and not the shortest distance. The principle links the ideas of ray optics and wave optics.
Fermat's principle states that the light between two points travels through the path where the optical length between two points is equal. It says that the path taken by light travelling between two points is the path that can be travelled in the shortest time.
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Derivation of the Laws of Refraction or Reflection
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Fermat's principle is used for the derivation of the laws of refraction or reflection. In the diagram given below, I is the angle of incidence and R is the angle of reflection.
Fermat's principle is used for the derivation of the laws of refraction or reflection. In the above diagram, a ray of light is shown where the angle of incidence and the angle of refraction are not equal. Therefore, we have to find out the time taken (t) by light to travel from point A to point B.
| t = ([ h12+ y2]+ [ h22+(w- y)2] ) / c . . . (1) |
where c is the speed of light. The minimum time is calculated by differentiating time t with respect to y and then the result is set to zero, the final result is,
| y / ([ h12+ y2]) = ( w - y ) / ( [ h22+(w- y)2]) . . . (2) |
Here, sin I = sin R
Therefore, it is seen that the left-hand side of the equation is sin I and the right-hand side of the equation is sin R . Hence, the minimum time for light to travel is reduced to
sin I = sin R
Or, I = R
which is known as the law of reflection.
Derivation of Snell's Law using Fermat's PrincipleSnell's law can also be derived using the statement and idea of Fermat's principle. Fermat's principle can also be consulted to derive Snell's law of refraction. In the above diagram, the area which is shaded blue is a medium with a refractive index of n > 1. Thus, the speed of light in a medium with refractive index n is c divided by n. Here, c is the speed of light in a vacuum.  Derivation of Snell's Law using Fermat's Principle On further solving, the result obtained from this equation is, ⇒ sin I = sin R Where R is known as the angle of refraction. This equation is known as Snell's law of refraction. |
Applications of Fermat’s Principle
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The applications of Fermat's principle are widespread. It links the ideas of ray optics and wave optics. Some applications are mentioned below:
- Fermat's principle brings forth that the angle of incidence of light on a surface is equal to that of the angle of reflection. This is also known as the Law of Reflection.
- Fermat’s principle states that the light rays are rectilinear in a homogeneous medium. It says that light travels in a straight line in any medium where the index of refraction is constant.
- This is one of the most fundamental principles of optics and is used in the derivation of a number of principles in geometric optics.
Examples of Fermat’s Principle
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Few examples of Fermat's principle in real life are as follows:
- Example 1: When we actually see the sun setting, it is already below the horizon. It does not look as though it is below the horizon, but it is. This is because the earth’s atmosphere is thin at the top and it is quite dense at the bottom. Light travels slower in the air than it does in a vacuum, and so the light of the sun can get to point beyond the horizon more quickly if, instead of just going in a straight line, it avoids the dense regions where it goes slowly by getting through them at a steeper tilt. When it appears to go below the horizon, it is actually already much below the horizon.
- Example 2: Another example of this phenomenon is the mirage we often see while driving on roads. We see water on the road, but when we get there, the road is as dry as the desert. The phenomenon occurs due to the following reason. What we are really seeing is the light of the sky which is being reflected on the road. Since the air is very hot just above the road but it is cooler up higher. Hot air expands more than cool air and is thinner, this leads to less decrease in the speed of light.
Solved MCQ Example Related to Fermat’s PrincipleQues. Assuming that a monochromatic light passes through 2 cm of glass (or 2.25 cm of water), then the optical path remains the same. Considering that the refractive index of glass is 1.50, determine the refractive index of water? (2 marks)
Ans. The answer is “a”, 1.33. The Refractive index product of the medium and its Geometric Distance travelled by light in the medium is called Optical Path. |
Previous Year Questions
- Reflection of the point … [KCET 2017]
- A and B are two parallel-sided transparent slabs of refractive indices … [WBJEE 2016]
- A ray of light passes through four transparent media with refractive indices … [JEE Advanced 2001]
- A transparent slab of thickness dd has a refractive index n(z) … [JEE Advanced 2016]
- The angle of minimum deviation for an incident light ray on an equilateral prism … [KCET 2009]
- In refraction, light waves are bent on passing from one medium … [KCET 2021]
- Two plane wavefronts of light, one incident on a thin convex lens … [KEAM]
- For a crystal, the angle of diffraction … [BITSAT 2008]
- When light is incident on a diffraction grating the zero-order … [KCET 2004]
- The width of the diffraction band varies … [KCET 2006]
- Assume that light of wavelength 600nm is coming from a star … [NEET 2020]
- An object is placed at a distance of 40 cm … [NEET 2018]
- The angle of a prism is A. One of its refracting surfaces is silvered. Light rays falling at an angle…? [NEET 2014]
- A cord attached to a vibrating tuning fork is divided into six segments under tension…? [VITEEE 2003]
- White light is incident on the interface of glass and air as shown in the figure. If green light is…? [JEE Mains 2004]
- Two plane mirrors A and B are aligned parallel to each other, as shown in the figure. A light ray is incident…? [JKCET 2002]
- A ray of light from a denser medium strikes a rarer medium at an angle of incidence I… [JEE Mains1983]
Things to Remember
- Fermat's principle is a part of ray optics that deals with the travelling of light between two points in such a way that the number of waves, which is the optical length between the points, is equal to that of the neighbouring path.
- Fermat’s principle is consistent with light travelling in a straight line in a homogeneous medium since it is rectilinear.
- Fermat's principle is used for the derivation of the laws of refraction or reflection.
- It was developed to find the path travelled by light in the shortest time, that is why Fermat's principle is also known as the principle of least time.
- The principle provides a link between the ideas of ray optics and wave optics.
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Sample Questions
Ques. What is the Basis of Fermat’s Law? [2 marks]
Ans. Fermat's principle is based on the concept that the angle of incidence and the angle of refraction of a light ray is equal. Light travelling through a straight line in a homogeneous medium is considered to be rectilinear. This principle states that light from a certain point A to another point B takes the least time. It also lays the groundwork for Snell's law which can be derived using this principle.
Ques: Describe the use of Fermat’s principle to derive Snell’s law. [3 marks]
Ans. Fermat's principle can also be consulted to derive Snell's law of refraction. In the above diagram, the area which is shaded blue is a medium with a refractive index of n > 1.
Thus, the speed of light in a medium with refractive index n is c divided by n. Here, c is the speed of light in a vacuum. The time required for the light ray to travel from point A to B in this medium will be n times the time that is required for the ray of light to travel the same distance in a vacuum. Therefore, the time required to go from point A to point B according to the diagram is,
t = ([ h12+ y2]+ n[ h22+(w- y)2] ) / c
On further solving, the result obtained from this equation is,
sin I = sin R
where R is known as the angle of refraction. This equation is known as Snell's law of refraction.
Hence, the law of reflection and Snell's law of refraction can be derived from Fermat's principle.
Ques. In Young’s double-slit experiment, the slits are repeated at 0.24 mm. The screen is 1.2 m away from the slits. The fringe width is 0.03cm. Calculate the wavelength of light used in the experiment? [3 marks]
Ans. The data provided -
β =0.3 cm= 3.0 x 10-3 m
D=1.2 m
d = 0.24 mm = 2.4 x 10-4 m
Now, β = Dλ/d
λ = βd/D
Substituting the values,
λ= ( 3.0 x 10-3 m x 2.4 x 10-4 m ) / 1.2 m
λ= 6.2 x 10-7 m.
Therefore, the wavelength of the light is found out to be 6.2 x 10-7 m.
Ques: Two coherent sources whose intensity ratio is 81:1 produce interference fringes. Calculate the ratio of the intensity of maxima and minima in the interference pattern? [3 marks]
Ans. Let I1 and I2 be two intensities,
The ratio is
I1 / I2 = 81 / 1
We know, Intensity ∝ ( Amplitude )2
a1/a2 = (81 / 1) = 9/1 = r
I max / I min = (r+1)2 / (r-1)2= ( 9+1/ 9-1 )2 = (10/8)2
I max / I min = 100/64 = 25/16
∴ I max : I min ratio is 25 : 16
Therefore, the ratios of the intensities of maxima and minima in the interference pattern are 25:16.
Ques. What is the polarization of light? What type of waves shows the property of polarization? Name any two methods to produce plane-polarized light. [3 marks]
Ans. Polarization of light is the phenomenon by which light waves are restricted to a certain direction that is perpendicular to the direction of the propagation of light. Transverse waves show the property of polarisation.
Two methods to produce plane-polarized light are:
- Polarization by Reflection
- Polarization by Scattering.
Ques: What is the shape of the wavefront in each of the following cases?
(a) Light diverging from a point source.
(b) Light emerges out of a convex lens when a point source is placed at its focus.
(c) The portion of the wavefront of the light from a distant star was intercepted by the Earth. [3 marks]
Ans. The shape of the wavefront in these cases are -
- Light diverging from a point source - spherical.
- Light emerges out of a convex lens when a point source is placed at its focus - a parallel grid.
- The portion of the wavefront of the light from a distant star was intercepted by the Earth - plane.
Ques. Monochromatic light of wavelength 589 nm is incident from the air on a water surface. What are the wavelength, frequency, and speed of: (a) Reflected light, and (b) refracted light? (Refractive index of water is 1.33) [5 marks]
Ans. (a) Reflected light
The data provided,
Wavelength of monochromatic light, λ = 589 nm = 589 x 10-9 m.
Speed of light in air, c = 3 x 10-8 m
Refractive index of water, μ = 1.33
In the given problem, the reflected ray will have the same speed, wavelength, and frequency as the incident ray. This is because the reflected ray will be reflected back to the same medium as the incident ray.
Frequency of light,
v=c/λ
v = ( 3 x 10-8 ) / ( 5000 x 10-10 ) = 6 ×1014 Hz
∠i + ∠r = 90°
∠i + ∠i = 90°
Zf= (4 x 10-3)2 / (400 x 10-9) = 40 m
or, ( λ1 - λ ) = 15 Å = 15 x 10-10 m
or, ( λ1 - λ ) = (v/c)λ
or, v = (c/λ) ( λ1 - λ )
Moreover, v/c = sin i/ sin r = μ
or, θ = 0.1o = 0.1 x λ/180 = 3.14/1800 rad
d = λ/θ
or, d = (600 x 10-9) / (3.14/1800) = 3.44 x 10-4 m
or, λ = a2/Zf
Now,
nλ = xd/D
or, x = nλD/d
or, θ = (d/d)λ)/d = λ/d
or, θ = (3 x 108)/(589x 10-9)
Therefore, θ = 5.09 x 1014 Hz
Hence, the speed, frequency and wavelength of reflected light are 3 x 108m,5.09 x 1014 Hz, 589 nm respectively.
(b) Refracted light.
Since the frequency of light does not depend on the medium-light is travelling in, the frequency of refracted rays in water will be equal to the frequency of the incident or reflected light in air. Therefore, the frequency of refracted light is 5.09 x 1014 Hz.
Speed of light in water is,
V = c/μ
or, V = (3×108)/1.33 = 2.26 x 108 m/s
Wavelength of light in water is,
λ = V/v
or, λ= (2.26 x 108)/(5.09 x 1014) = 444.007 x 10-9 m
Therefore, λ = 444.01 nm.
Hence, the speed, frequency, and wavelength of refracted light are 2.26 x 108 m/s, 444.01 nm, 5.09 x 1014 Hz respectively.
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