Topic : Define prism , the angle of prism , Refraction Through Prism ,Prism Formula ( Condition for Minimum Deviation
Define Prism and the angle of prism :
A simple prism is a homogeneous transparent refracting medium bounded by at least two non parallel plane surfaces inclined at some angle.
The surface on which light is incident and another surface from which light comes out should be non parallel.The two non parallel plane surfaces participating in refraction of light are called refracting surfaces and the line of intersection of the two refracting surfaces is called refracting edge.
The angle between two refracting surfaces is called the angle of prism or refracting angle and is denoted by A.
Refraction Through Prism:
Discuss refraction of light due to a prism and hence show that $\delta + A = i + e$,
Determination of angle of deviation ($\delta$).
Let ABC be the principal section of a prism of refracting angle A. Let a ray of light DE be incident on the refracting surface AB of the prism at an angle of incidence $i$. After refraction at E, the ray of light bends towards the normal NO and travels along EF. The refracted ray EF again suffers a refraction at F and bends away from the normal N'O and travels along FG. The ray FG is called emergent ray. The angle made by the emergent ray with the normal is called angle of emergence ($e$). When the emergent ray is produced backward, it meets the incident ray produced forward at point O'. The angle between the emergent ray and the incident ray is called angle of deviation ($\delta$).
The incident ray DE is deviated along EF at surface AB of the prism.
$\angle HEF = \angle HEO - \angle FEO$
$\delta_1 = i - r_1.....(i)$
where $\delta_1$ is the deviation produced by the surface AB of the prism.
Similarly, at surface AC, the ray EF is deviated along FG, so
$\delta_2 = e - r_2.....(ii)$
where $\delta_2$ is the deviation produced by the AC surface of the prism.
$\delta = \delta_1 + \delta_2$
$\delta = i - r_1 + e - r_2/$
$\delta= (i + e) - (r_1 + r_2).....(iii)$
From quadrilateral AEOF,
$\angle A + \angle AEO + \angle EOF + \angle OFA = 360^\circ$
But $\angle AEO = \angle OFA = 90^\circ$
$\angle A + 90^\circ + \angle EOF + 90^\circ = 360^\circ$
$\angle A + \angle EOF = 180^\circ....(iv)$
From $\triangle EOF$, $\angle r_1 + \angle EOF + \angle r_2 = 180^\circ.....(v)$
From eqns. (iv) and (v), we get
$\angle A + \angle EOF =\angle r_1 + \angle EOF + \angle r_2$
$A = r_1 + r_2.....(vi)$
Substituting the value of eqn. (vi) in eqn. (iii), we get,
$\delta = (i + e) - A$
$\delta + A = i + e$
Prism Formula (Angle of minimum deviation and refractive index of a prism)
Question : Show the variation of angle of deviation produced by a prism with the angle of incidence and hence define angle of minimum deviation. ?
The angle of deviation depends on the angle of incidence ($\therefore \delta = (i + e) - A$).
The variation of angle of deviation ($\delta$) with angle of incidence for a triangular prism is shown in figure :
As the angle of incidence of the light increases, the angle of deviation ($\delta$) decreases till it becomes minimum at a particular angle of incidence. The minimum value of the angle of deviation is called the angle of minimum deviation ($\delta_m$). The position of the prism at a minimum angle of deviation is known as minimum deviation position.
Condition for Minimum Deviation :
Question: State the condition for minimum deviation.
In the position of minimum deviation, the prism lies symmetrically with respect to the incident ray and the emergent ray i.e., $i = e$.
Proof :
When the light ray incident on the face AB of the prism and refracted into the prism. we have
$ {}_{a}n_g = \frac{\sin i}{\sin r_1} \quad \cdots (i) $
When the refracted ray incident on the face AC of the prism and emerges out into the air, we have,
$ {}_{g}n_a = \frac{\sin r_2}{\sin e} $
$ {}_{g}n_a = \frac{1}{{}_{a}n_g} $
$\therefore \quad {}_{a}n_g = \frac{\sin e}{\sin r_2} \quad \cdots (ii) $
From eqns. (i) and (ii), we get,
$ \frac{\sin i}{\sin r_1} = \frac{\sin e}{\sin r_2} \quad \cdots (iii) $
$ i = e$ and $r_1 = r_2 $
Therefore, we conclude that when the prism is in the minimum deviation position, the refracted ray inside the prism is parallel to the base of the prism and $i = e$ and $r_1 = r_2$.
Prism Formula - Refractive Index of the Material of a Prism
Question: Find the expression for the refractive index of the material of a prism. Hence calculate the deviation produced by a prism of small angle.
For a prism
$\delta + A = i + e, \quad \cdots (i)$
$r_1 + r_2 = A \quad \cdots (ii)$
If deviation, $\delta = \delta_m; i = e$
and $r_1 = r_2 = r$
Then, from eqn. (i), we have
$\delta_m + A = 2i$
$i = \frac{A + \delta_m}{2} \quad \cdots (iii)$
And from equation (ii),
$2r = A$
$r = A/2 \quad \cdots (iv)$
According to Snell's law,
$n = \frac{\sin i}{\sin r}$
Substituting the values of 'i' and 'r' from eqns. (iii) and (iv), we get,
$n = \frac{\sin[(A + \delta_m)/2]}{\sin(A/2)} \quad \cdots (v)$
Eqn. (v) is known as prism formula.
For a thin prism, or small angle prism, A and $\delta_m$ both are very small, so equation. (v) can be written as
$n = \frac{(A + \delta_m)/2}{A/2}$
Thus, deviation produced by a prism of small angle is given by,
$\delta_m = (n - 1)A \quad \cdots (vi)$
Note :
1. Angle of minimum deviation $\delta_m$ is also sometimes denoted as $D_m$
2. Deviation produced by a very thin prism does not depend upon angle of incidence i.e., it is a constant quantity for a given prism.
3. A thin prism gives less deviation of light.