Question

Relation between refractive index and angle of minimum deviation

Answer 1

As a ray of light enters any medium, the ray's direction is deflected, based on the entrance angle (typically measured relative to the surface normal), the material's refractive index, and according to Snell's law. A beam passing through an object like a prism or water drop is deflected twice: once entering, and again when exiting. The sum of these two deflections is called the deviation angle.

The deviation angle in a prism depends upon:

Refractive index of the prism: The refractive index depends on the material and the wavelength of the light. The larger the refractive index, the larger the deviation angle.

Angle of the prism: The larger the prism angle, the larger the deviation angle.

Angle of incidence: The angle at which the beam enters the object is called the angle of incidence. The deviation angle first increases with increasing incidence angle, and then decreases.

There is an angle of incidence at which the sum of the two deflections is minimal. The deviation angle at this point is called the "minimum deviation" angle, or "angle of minimum deviation".[1] At the minimum deviation angle, the incidence and exit angles of the ray are identical. One of the factors that causes a rainbow is the bunching of light rays at the minimum deviation angle that is close to the rainbow angle.

A convenient way to measure the refractive index of a prism is to direct a light ray through the prism such that it produces the minimum deviation angle. This yields a simple formula:[2]

{\displaystyle n_{\lambda }={\frac {\sin({\frac {A+D_{\lambda }}{2}})}{\sin({\frac {A}{2}})}},} {\displaystyle n_{\lambda }={\frac {\sin({\frac {A+D_{\lambda }}{2}})}{\sin({\frac {A}{2}})}},}

where {\textstyle n_{\lambda }} {\textstyle n_{\lambda }}is the refractive index at a wavelength {\displaystyle \lambda } \lambda , {\displaystyle D_{\lambda }} D_{\lambda } is the angle of minimum deviation, and {\displaystyle A} A is the internal angle of the prism.

The angle of minimum deviation is attained when the angle of incidence and angle of emergence for a ray of light being refracted through a prism, are equal.

Also, the variation of angle of deviation with an arbitrary angle of incidence can be encapsulated into a single equation:

{\displaystyle D=i-A+\arcsin \left(n\cdot \sin \left(A-\arcsin \left({\frac {\sin i}{n}}\right)\right)\right),} {\displaystyle D=i-A+\arcsin \left(n\cdot \sin \left(A-\arcsin \left({\frac {\sin i}{n}}\right)\right)\right),}

where {\displaystyle D} D is the angle of deviation at an arbitrary angle of incidence {\displaystyle i} i, and {\displaystyle n} n is the refractive index of the prism.

It may be noted that minimum deviation in a rectangular slab is zero, as both the surfaces separating two media are parallel, whereas in a prism they are inclined to each other.

Answer 2

To Find :

Relation between refractive index and angle of minimum deviation .

Solution :

Let , A be apex angle and $\delta_m$ is minimum deviation .

We know , at $\delta_m$ , i = e and $r_1=r_2=r\ ( say )$ .

Now ,

$A=r_1+r_2 \\A =2r\\r=\dfrac{A}{2}$

Also , $i+e=A+\delta$

At minimum deviation .

$2i=A+\delta_m\\i=\dfrac{A+\delta_m}{2}$

Now , according to Snell's law :

$\mu=\dfrac{sin \ i}{sin\ r}$

Putting value of  i and r in above equation :

$\mu=\dfrac{sin \dfrac{(A+\delta_m)}{2}}{sin\dfrac{A}{2}}$

Therefore , this is the required solution .

Learn More :

Refraction

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