Laws of Refraction of Light, Relation between Relative Refractive Indices and Lateral Shift

Light travels along a straight line as it moves in one medium. But when it moves from one medium to another it bends. The phenomenon of bending of light as it moves from one medium to another is called refraction of light.

Consider a surface XY which separates the two media air and glass. When a ray of light OA is incident at point A on the surface, it bends towards the normal along AB. Here NN’ is a normal drawn at point A as shown in the figure. Here OA is incident ray, AB is refracted ray, ∠NAB is the angle of refraction.

Air is a rarer medium and glass a denser medium. Those substances which have more density are called denser and which has less density are called rarer medium.

Laws of Refraction of Light

The laws of refraction of light are:

1. The incident ray, refracted ray and normal at the point of the incident all lie in the same plane.
2. When light travels from rarer medium to denser medium, it bends towards the normal and when light travels from denser medium to rarer medium, it bends away from normal.
3. For given two media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant, called refractive index it is denoted by µ.
   $$ _a\mu _b = \frac {\sin i}{\sin r} $$
   where _aµ_b is the refractive index of medium b with respect to medium a. when light travels through the vacuum to any other medium, the refractive index is called absolute refractive index. The ratio of the velocity of light is the vacuum to the velocity of light in that medium is defined as the refractive index of a medium.
   \begin{align*} \text {absolute refractive index}, \mu &= \frac {\text {velocity of light in vacuum}}{\text {velocity of light in vacuum}}\\\text {velocity of light in medium} &= \frac cv \end{align*}

Relation between Relative Refractive Indices

Suppose a ray of light AO in air is refracted into water along OB> the refractive index of water is

$$_a\mu _w = \frac {\sin i}{\sin r} $$

If the refracted ray OB reverses its path to BO, as shown in the figure, the ray will retrace the initial path OA as before. This is called the principle of reversibility of light. So BO is now the refraction equal to i. BO is now the incident ray with the angle of incidence equal to R and OA the refracted ray in the air with an angle of refraction equal to i. So,

\begin{align*} _w\mu _a = \frac {\sin r}{\sin i} \\ &= \frac {1}{(\sin i/\sin r)} \\ &= \frac {1}{_a \mu _w} \\ _a\mu _w &= \frac {1}{_w\mu _a} \end{align*}

Suppose three media: air, water, and glass separated each other by parallel surfaces, as shown in the figure. A ray of light AO incident on air-water surface is refracted along OB in water and again BC in glass and emerges out along CD in air. Since the surfaces separating the media are parallel, the final emergent ray will be parallel to incident ray AO, and then the angle of emergence is equal to the angle of incidence.

Let i_a and i_w be the angle of incidence and angle of refraction for ray OA and OB respectively. Then, the relative refractive index,

\begin{align*} _a\mu _w &= \frac {\sin i}{\sin r} = \frac {\sin i_a}{\sin i_w} \\ \text {From water to glass,} \\ _w\mu _g &= \frac {\sin i_w}{\sin i_g} \\ \text {and from glass to air,} \\ _g\mu _a &= \frac {\sin i_g}{\sin i_a} \\ \text {Now, multiplying three refractive indices, we get} \\ _a\mu _w \times _w \mu _g \times _g\mu _a &= {\sin i_a}{\sin i_w}\times {\sin i_w}{\sin i_g} \times {\sin i_g}{\sin i_a} \\ \text {or,} \:_a\mu _w \times _w \mu _g \times _g\mu _a &= 1 \\ \text {From this equation, we have} \\ _w\mu _g &= \frac {1}{_a\mu _w \times _g\mu _a} = \frac {_a\mu _g}{_a\mu _w} \\ \end{align*}

The refractive index of water and glass are 1.33 and 1.5 respectively. So, the relative index of glass with respect to water is 1.5/1.33 = 1.12.

\begin{align*} \text {Again, we have} \: \frac {\sin i}{\sin r} = _a\mu _w, \text {for air water surface} \\ \text {or,} \: \sin i_a &= _a\mu _w \sin i_w \\ \text {and} \: \frac {\sin i_a}{\sin r_g} &= _a\mu _a, \text {for air glass surface} \\ \text {or,} \: \sin i_a &= _a\mu _g \sin i_g \\ \text {Then, we have} \\ \sin i_a &= _a\mu _w \sin i_w = _a\mu _g \sin i_g \\ \text {In terms of absolute refractive index,} \\ \mu _a \sin i_a &= \mu _w\sin i_w = \mu _g \sin i_g \\ \end{align*}

This relation shows that if a ray of light is refracted from one medium to another with parallel boundaries,

$$ \mu \sin i = \text {constant} $$

Where µ is the absolute refractive index of a medium and I is the angle made by the ray with the normal in that medium.

Lateral Shift

When a ray of light is incident on the glass slab, the emergent ray coming out of the slab will be parallel to the incident ray. The perpendicular distance between the direction of incident ray and emergent ray called the lateral shift. It is denoted by d.

Consider a glass slab MNPS of thickness t. When a ray of light OA is incident on the surface of the slab at A, it is refracted along AB and at last emerges out along BC is shown in the figure. N₁N₂ and N₃N₄ be the incident ray and emergent ray and BQ is lateral shift.

\begin{align*} \text {From} \Delta ABQ, \text {we have} \\ \sin (i-r) &= \frac {BQ}{AB} \\ \text {or,} \: BQ &= AB \sin (i-r) \dots (i) \\ \text {Again from} \Delta ABN_2, \text {we have} \\ \cos r &= \frac {AN_2}{AB} = \frac {t}{AB} \\ \text {or,} \: AB &= \frac {t}{\cos r} \\ \text {So equation}(i) \text {becomes} \\ BQ &= \frac {t}{\cos r} \sin (i-r) \\ \therefore BQ &= \frac {t \sin (i-r)}{\cos r} \\ \text {Hence, when} I = 90^o \\ BQ &= d = \frac {t \sin (90^o-r)}{\cos r}\\ &= \frac {t\cos r}{\cos r} \\\therefore d &= t \end{align*}

It means that when light is incident at 90^o on the surface of glass slab, lateral shift produced by it is equal to the thickness of the slab.