Snell's Law

Snell's law, also known as the law of refraction, is a law stating the relationship between the angles of incidence and refraction, when referring to light passing from one medium to another medium such as air to water, glass to air, etc.

Snell's Law states that the ratio of sine of angle of incidence and sine of angle of refraction is always constant for a given pair of media.

$$\dfrac{\sin i}{\sin r}=\text{constant}=n=\text{refractive index}$$

Let us consider that light enters from medium 1 to medium 2,

$$\therefore \dfrac{\sin i}{\sin r}=n_{21}=\dfrac{n_2}{n_1}=\dfrac{\color{#3D99F6}{v_1}}{\color{#3D99F6}{v_2}}=\dfrac{\color{#3D99F6}{\lambda_1}}{\color{#3D99F6}{\lambda_2}}$$

Here, $v_n$ is the velocity of light in respective medium and $\lambda_n$ is the wavelength of light in respective medium. You may be wondering how we obtained the expression in blue color, well if we define it in an easy way, the basic cause of refraction is due to the change in velocity of light by entering a medium of different refractive index. So, if a medium has less refractive index, then the velocity of light in that medium would be more but if a medium has more refractive index then the velocity of light in that medium would be comparatively less.

$$\therefore v \propto \dfrac{1}{n} \Rightarrow \dfrac{v_1}{v_2}=\dfrac{n_2}{n_1}=n_{21}$$

Question: A ray of light travelling in air is incident on the plane surface of a transparent medium. The angle of incidence is found to be $45^{\circ}$ and the angle of refraction is $30^{\circ}$. Find the refractive index of the medium.

---

Solution: We know that $\hat i=45^{\circ}$ and $\hat r=30^{\circ}$

Therefore refractive index, $$\begin{aligned} n=\dfrac{\sin i}{\sin r} &= \dfrac{\sin 45^{\circ}}{\sin 30^{\circ}}\\ &= \dfrac{1/\sqrt{2}}{1/2}= \sqrt{2} \end{aligned}$$

Absolute Refractive Index:

When we compare the speed of light in a medium to that of the speed of the light in vacuum, then we would be dealing with something called absolute refractive index. We generally refer to the absolute refractive index of a medium when we say that a certain object's refractive index is $x$.

The expression for the absolute refractive index of a medium would thus be: $$\text{absolute refractive index}=\dfrac{\text{speed of light in vacuum}}{\text{speed of light in the given medium}} = \dfrac{c}{v}$$

Note: As the speed of light is at its maximum in vacuum, the absolute refractive index always greater than $1$. Also note that the refractive index is a relative quantity and thus it had no units.

Question: The absolute refractive index of a glass window is $1.5$. What is the speed of light when it is traveling through the glass window? Assume that the speed of light in vacuum $=3\times 10^8m/s$.

---

Solution: According to the question, we have: $$\dfrac{\text{speed of light in vacuum}}{\text{speed of light in the given medium}}=1.5\\ \implies \dfrac{3\times 10^8}{\text{speed of light in the given medium}}=1.5\\ \implies \text{speed of light in the given medium}=\dfrac{3\times 10^8}{1.5}=\boxed{2\times 10^8 m/s}$$

Question: The absolute refractive index of diamond is $2.42$. What is the speed of light in diamond? (Take speed of light in vacuum= $3 \times 10^8 m/s$

---

Solution: Absolute refractive index of diamond is $$=\dfrac{\text{speed of light in vacuum}}{\text{speed of light in diamond}}\quad\therefore\dfrac{c}{v}=2.42\\ \implies v=\dfrac{c}{2.42} \implies v=\dfrac{3 \times 10^8}{2.42} \\\boxed{v=1.24 \times 10^8 m/s}$$

Refraction of a ray of light in a glass slab

In this case, we will try to prove $\angle i_1=\angle r_2$ or the incident ray is parallel to the emergent ray,

Applying Snell's Law when the light is incident on the glass slab's surface,

$$\dfrac{\sin i_1}{\sin r_1}=n=\text{refractive index of glass}$$

Now, applying Snell's Law when the light ray is leaving the glass slab through another surface,

$$\dfrac{\sin i_2}{\sin r_2}=\dfrac{1}{n}\Rightarrow \dfrac{\sin r_2}{\sin i_2}=n=\text{refractive index of glass} \\ \therefore \dfrac{\sin i_1}{\sin r_1}=\dfrac{\sin r_2}{\sin i_2}$$

Now, $\angle r_1=\angle i_2$ as they are alternate angles, thus, $\sin r_1=\sin i_2$,

$$\therefore \sin i_1=\sin r_2\Rightarrow \angle i_1=\angle r_2$$

So, the incident ray is parallel to the emergent ray but it is laterally displaced from it.

Question: A ray of light travelling in air falls on the surface of a transparent glass slab. The ray makes and angle of $45^{\circ}$ with the normal to the surface. Find the angle made by the refracted ray with the normal within the slab. Given that refractive index of the glass slab is $\sqrt{2}$.

---

Solution: We know that $n=\dfrac{\sin i}{\sin r} = \dfrac{\sin 45^{\circ}}{\sin r}$, here the refractive index is $\sqrt{2}$.

$$\begin{aligned} \dfrac{\sin 45^{\circ}}{\sin r}&=\sqrt{2}\\ \implies\sin r &= \dfrac{1}{\sqrt{2}}\times \sin 45^{\circ}\\ =\dfrac{1}{\sqrt{2}}\times \dfrac{1}{\sqrt{2}} &=\dfrac{1}{2} \end{aligned}$$

Thus, as $\sin r$ = $\dfrac{1}{2}$, the angle of refraction would be $r=\sin^{-1}\left(\dfrac 12\right)=30^\circ$.

As discussed earlier, the emergent ray is parallel to the incident ray but appears slightly shifted, and this shift in the position of the emergent ray as compared to the incident ray is called Lateral displacement.

Lateral Displacement

The perpendicular distance between the incident ray and the emergent ray is defined as lateral shift. This shift depends upon the angle of incidence, the angle of refraction and the thickness of the medium. It is given by the following expression: $$S_{\text{Lateral}}=\dfrac{t}{\cos r}\sin{(i-r)}$$

We shall now try to derive the above stated formula for a Glass slab. In the figure given below, $AB$ is the incident ray, $BC$ is the refracted ray and $CD$ is the emergent ray. The ray is striking the slab at an angle of $i_1$ and it is emerging from the slab at an angle of $r_2$.

Refraction of a ray of light in a glass slab with it's corresponding angles

In $\triangle BCK$,

$$\sin (i_1-r_1)=\dfrac{CK}{BC} \Rightarrow CK=BC \sin (i_1-r_1)$$

In $\triangle BCN'$,

$$\cos r_1=\dfrac{BN'}{BC}=\dfrac{t}{BC} \Rightarrow BC=\dfrac{t}{\cos r_1}$$

Here, $t$ is the thickness of slab.

Substituting the value of $BC$ in the first equation,

$$S_L=\text{Lateral Displacement }(CK)=t\dfrac{\sin(i_1-r_1)}{\cos r_1}$$

Question: The thickness of a glass slab is $0.25m$, it has a refractive index of $1.5$. A ray of light is incident on the surface of the slab at an angle of $60^\circ$.

Find the lateral displacement of the light ray when it emerges from the other side of the mirror. You may assume that the speed of light is $3\times 10^8 m/s$.

---

Solution: From the previous topics, we know: $$\text{refractive index}=\dfrac{\sin i}{\sin r}=1.5\text{ (in this case)}\\\sin r=\dfrac{1.5}{\sin 60}\approx 0.57735\\\implies r = \sin^{-1}(0.57735)\approx 35.25^\circ$$

Now, applying the values in the formula for lateral displacement we get: $$S_L=\dfrac{0.25}{\cos(35.25)}\times\sin(60-35.25)\approx 0.1281 m =\boxed{12.81cm}$$

Many a time you might have seen the floor of the swimming pool raised/ the letters appearing to be raised under a glass slab, ever wondered why this happens? If you observe clearly, you'll find that refraction explains it. Let's see the definition.

The vertical distance by which an object appears to be shifted when an object placed in one medium is observed from another medium of different refractive indices, is called Normal shift. It is given by the formula:

\[S_{\text{Normal}}=t\left(1-\dfrac{1}{_{\text r}n_{\text d}}\right)\quad\text{where}\quad _{\text r}n_{\text d}=\mu=\dfrac{\text{real depth}}{\text{apparent depth}}\]

Note that: $$\text{refractive index}(\mu)=\dfrac{\sin i}{\sin r}=\dfrac{\text{real depth}}{\text{apparent depth}}$$

The thickness of a glass slab is $0.2m$, and it is placed over a flat book, the refractive index of the glass slab is $1.5$. A student looks through it and finds that the normal shift is $x$, find $x$.

---

Solution: We know that: $$\begin{aligned} S_N&=t\left(1-\dfrac{1}{\mu}\right)\\ &=0.2\left(1-\dfrac{1}{1.5}\right)\\ &=0.2\times \dfrac 13= 0.066m \end{aligned}$$

When light travels from a denser to rarer medium with an angle greater than the critical angle, the ray of light does not deviate in its path or does not refract, but it undergoes a reflection known as total internal reflection. The angle beyond which light in a given medium undergoes total internal reflection is called the critical angle.

The critical angle differs from medium to medium. If the refractive index of a given medium is $\mu$, then it's critical angle is given by the formula: [1]

$$\mu =\dfrac { 1 }{ \sin{ \theta }_{ c } }\quad\\\theta_c=\sin^{-1}\left(\dfrac 1\mu\right)$$

This is very useful as it is used in fiber glasses where total internal reflection helps in fast movement of wavelengths.

Sparkle of the diamond

Whenever your mom wears it you notice it, yes the sparkling beauty of the diamond never misses our eye. But have you ever wondered why the diamond sparkles? Well it is due to the phenomenon we've been discussing now, total internal reflection. Sparking beauty of the hope diamond [2]

Mirage Formation

This very old illusion ,which had fooled many people, is due to the magic of Total Internal relfection! Mirage is an optical illusion caused by refraction and total internal reflection. We know that the temperature of air varies with height, and also refractive index depends on the temperature of the medium.

Mirage Formation on a road [3]

During hot summers, the Surface of the Earth gets hotter, and the layers of air with decreasing temperature are formed. But the hot air has a refractive index lower than the cold air, that is hot air is optically rarer than cold air, and we know if a ray of light passes through a rarer medium from a denser medium, then the light rays bend away from the normal. So, at some points the light rays get totally reflected internally and reach the eyes of an observer, creating the reflection of an object on the surface of the Earth.

Looming

Very similar to mirage formation,thus phenomenon makes the objects appear to be levitating in the sky. This is mostly seen in the polar regions (as opposed to mirages, which are generally frowned in hot deserts). In these places the surface of the Earth is very cold and as we go up, layers of air with increasing temperatures are formed. As a result, the layers of atmosphere near the Earth have a higher refractive index than the layers above them, this layer is called as an inversion layer.

The objects appear to be floating due to the phenomemnon of looming [4]

When the light from any object (normally ships) reaches an observer, it undergoes a series of refractions which makes the light rays bend away from the normal, and at a point, they reach a stage where the angle incidence is greater than the critical angle and thus the rays undergo total internal reflection and reach the eye of an observer and creates and optical illusion that the object is really floating in the sky!

Fibre Optics

Optical fibres are the devices used to transfer light signals over large distances with negligible loss of energy. It is a revolutionary idea in terms of communication. But it's working is based on this simple phenomenon of total internal reflection. If you take a close look at an optical fibre you will observe that it consists of a thin transparent material, this is know as the core. This core is coated with something known as cladding and has a higher refractive index than the surrounding medium[5], it prevents the absorption of light by any means.

The internal structure and the transfer of light signal in a single optical fibre [6]

When the light rays enter the acceptance cone, some rays which are incident at an angle greater than the critical angle gets reflected internally and then it undergoes a series of Total Internal reflections until it reaches the other end of the firbe. But we should note that not all of the rays get reflected internally because they may not have struck the surface at the required angle (as seen in the figure above).

[1] Total Internal reflection, rp-encyclopedia.com. Retrieved 16:45, March 15, 2016, from https://www.rpphotonics.com/totalinternalreflection.html.

[2] Image from https://en.m.wikipedia.org/wiki/Diamond#/media/File%3ATheHopeDiamond-SIA.jpg under the creative Commons license for reuse and modification.

[3] Image credit http://epod.usra.edu/blog/2010/03/highway-mirage.html: Universities Space Research Association

[4] Image from https://en.m.wikipedia.org/wiki/File:Illustrationofloomingrefrationphenomenon.jpg under the creative Commons license for reuse and modification.

[5] Optical Fibres, rp-encyclopedia.com. Retrieved 08:56, March 17, 2016, from https://www.rpphotonics.com/fibers.html.

[6] Image credit http://www.pacificcable.com/Fiber-Optic-Tutorial.html.