Name: Total Internal Reflection From Prism
Uploaded: 2019-07-19
Duration: 1 min 56 s
Description: You must have seen reflectors on a bicycle. Why are they attached to the mudguards? What use do they have? Let’s watch this video on reflectors to find out.

Question

Derive an equation

δ = i + e - A

for an equilateral prism and using it, obtain an equation for the refractive index

(n)

of the material of the prism.

Accepted Answer

Let light ray $PQ$ be the incident ray. The angles of incidence and refraction at the first face $AB$ are $i_{1}$ and $r_{1}$ respectively. $QR$ is the path followed by the light inside the prism. The angle of incidence and refraction at the second face $AC$ are $r_{2}$ and $i_{2}$ respectively, $RS$ is the emerging ray from the second face. Angle $i_{2}$ is also called the angle of emergence $(e)$ . $δ$ is the angle of deviation formed between the direction of the incident ray $PQ$ and the emergent ray $RS$ . The two normals are drawn at the point of incidence $Q$ and emergence $R$ are $QN$ and $RN$ . They meet at the point $N$ . The incident ray and emergent Ray meet at a point $M$ .

The angle of deviation $δ_{1}$ at the surface $AB$ is, $∠ RQM = δ_{1} = i_{1} - r_{1}$ .

The angle of deviation $δ_{2}$ at the surface $AC$ is, $∠ QRM = δ_{2} = i_{2} - r_{2}$ .

The total angle of deviation $δ = δ_{1} + δ_{2}$ .

By substituting the values of $δ_{1} and δ_{2}$ , $δ = (i_{1} - r_{1}) + (i_{2} - r_{2})$ .

In the quadrilateral $AQNR$ two of the angles at the vertices $Q$ and $R$ are right angles therefore, the sum of the other angle of the quadrilateral is $180 °$ .

Hence, $∠ A + ∠ QNR = 180 ° - - - - - - - - - - (i)$ .

Now, from the triangle $QNR$ ,

$r_{1} + r_{2} + ∠ QNR = 180 ° - - - - - - - - - - (i i)$

Comparing equation $(i)$ and $(i i)$

$r_{1} + r_{2} = A$

Now, by substituting the obtained value in $δ = (i_{1} - r_{1}) + (i_{2} - r_{2})$ .

$δ = i_{1} + i_{2} - A$ . This is the angle of deviation for an equilateral prism.

As, $i_{2} = e$ (angle of emergence)

Therefore, it can be rewritten as $δ = i_{1} + e - A$ .

Now, let us obtain the refractive index of the material of the prism.

At minimum deviation, $i_{1} = i_{2} = i and r_{1} = r_{2} = r$ .

Now using the equation for the angle of deviation i.e., $δ = i_{1} + i_{2} - A = 2 i - A or i = (\frac{A + δ}{2})$ .

And, as $r_{1} + r_{2} = A = 2 r = A or r = \frac{A}{2}$ .

Now according to Snell's law $n = \frac{\sin i}{\sin r}$ .

Now substituting the value of $i and r$ in above expression.

$n = \frac{\sin (\frac{A + δ}{2})}{\sin \frac{A}{2}}$ .

Therefore, the refractive index of the material of the prism can be obtained by using this expression i.e., $n = \frac{\sin (\frac{A + δ}{2})}{\sin \frac{A}{2}}$

Derive an equation $δ = i + e - A$ for an equilateral prism and using it, obtain an equation for the refractive index $(n)$ of the material of the prism.

Important Questions on Ray Optics and Optical Instruments

Learn and Prepare for any exam you want !

Derive an equation δ=i+e-A for an equilateral prism and using it, obtain an equation for the refractive index (n) of the material of the prism.

Important Questions on Ray Optics and Optical Instruments

Learn and Prepare for any exam you want !

Derive an equation $δ = i + e - A$ for an equilateral prism and using it, obtain an equation for the refractive index $(n)$ of the material of the prism.