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Snell’s law tells about the degree of refraction and the relation between the angle of incidence, the angle of refraction, and the refractive indices of a pair of media.

As light travels from one medium to the other, it experiences refraction. Snell's law calculates the degree to which the light bends. Therefore, this law has been termed the law of refraction. In the year 1621, physicist Willebrord Snell discovered the law of refraction and hence the name Snell’s law.

According to Snell’s law - The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for the light of a given color and for the given pair of media.

Formula of Snell’s law – sin i / sin r = constant = µ.

Here 'i' is the angle of incidence, and 'r' is the angle of refraction. The constant value is called the refractive index of the second medium with respect to the first.

Diagrammatic representation is as follows-

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Snell's law is important because it is used to determine the direction of the light rays through refractive media with varying indices of refraction. For example, the indices of the refraction of both the media in the above picture named n_{1 }and n_{2 }are used to represent the factor by which the light ray's speed keeps decreasing when passing through a refractive medium, as opposed to its velocity in a vacuum.

When a ray of light passes through the border between two media, the light will either be refracted to a lesser angle or a greater one. The situation is entirely dependent on the relative refractive indices of the two media in the scene. These angles are measured concerning the *normal *(a perpendicular line to the boundary).

Refraction between two media can also be reversible. If and only if all conditions are identical, the angles would be the same for light traveling in the opposite direction.

Physicists say Snell's law is only true for isotropic or specular media. In the case of any anisotropic media, birefringence may split the refracted ray into two rays – the o-ray or the ordinary ray, which follows Snell's law, and the e-ray or the extraordinary ray, which might not be co-planar with the incident ray.

Now when the light is monochromatic, that is, of a single frequency, Snell's law can also be expressed in terms of a ratio of wavelengths in the two different media as y_{1} and y_{2}.

Sin i / Sin r = v_{1}/v_{2 = }y_{1}/y_{2.}

To understand Snell’s Law, you also have to understand in what situations is the law applicable. This is because Snell's law applies to the refraction of light in any and every situation, irrespective of what the two media are.

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*Example 1 – *

*Question - *Light travels from air into an optical fiber with an index of refraction of 1.44. (a) In which direction does the light bend? (b) If the angle of incidence on the end of the fiber is 22^{o}, what is the angle of refraction inside the fiber? (c) Sketch the path of light as it changes media.

*Solution - *Since the light travels from a rarer region (lower n) to a denser region (higher n), it will bend toward the normal.

We will identify air as medium 1 and fiber as medium 2. Thus, n1 = 1.00, n2 = 1.44, and θ/font>1 = 22o. Snell's law then becomes

(1.00) sin 22o = 1.44 sin θ2.

sin θ2 = (1.00/1.44) sin 22o = 0.260

θ2 = sin-1 (0.260) = 15o.

The path of the light is shown in the figure below:

*Example 2 – *

*Question***:** If a ray is refracted at an angle of 14° and the refractive index is 1.2, compute the angle of incidence.

*Solution:*** **Given,

The angle of refraction, r = 14°

Refractive index, μ = 1.2

Using Snell’s law formula,

Sin i / Sin r = µ

Sin i / sin 14 = 1.2

Sin i = 1.2 * Sin 14

- 1,2*0.4
- 28

i = sin ^{-1} (0.288) = 16.7

*Example 3 – *

* Question *– A cup of water is placed on a table outside in the sunlight and is filled to the top. The sun happens to be 45o above the horizon, and you notice the bottom of the cup has just become completely shaded. The cup has a depth of 15cm, and you wanted to calculate the width of the bottom of the cup. Find the width in cm.

*Solution –*

Let α be the angle w.r.t. the horizon that the sun is at, i.e. 45, and d is the depth of the water.

Since the light passes from air into water n2= 1.33; n1 = 1.θ1 + α = 90°θ1 = 90° – 45° = 45°

From Snell’s law, we know that Sin(θ1)/ Sin(θ2) = n2/n1 Sin(45°)/ Sin(θ2) = 1.33/1Sin(θ2)= Sin(45°)/1.33 = 0.53 θ2= Sin-1(0.53)= 32.1°

d and w form a right-angle triangle with an angle of θ2.

Using the tangent tan(θ2) = w/dw = d tan(θ2) = 15 tan (32.1°) w = 9.4cm

Width of the cup = 9.4cm.

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The law has huge applications in the field of optics. For example, it is used in ray tracing to find and compute the angle of indices of refraction. It is also used in experimental optics to find the refractive index of a material. The law is also applicable in the case of meta-materials, which permits light to bend or refract backward, giving a negative angle of refraction with a negative refractive index.

If real-life applications are considered, Snell's law has a wide range of applications in Physics. For example, it is used in different optical apparatuses like eyeglasses, contact lenses, and cameras.

A device called a refractometer uses Snell's law widely. For example, the device uses the law to calculate the refractive index of liquids.

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