Ask Questions, Get Answers

Questions  >>  CBSE XII  >>  Physics  >>  Wave optics

<div>What is dispersion?What is the angle of deviation?Derive an expression for minimum angle of dev

$\begin{array}{1 1} (a)\;\mu= \sin \bigg(\large\frac{A+D_m/2}{A/2}\bigg)\\ (b)\;\mu= \bigg(\large\frac{D_m/2}{A/2}\bigg) \\ (c)\;\mu= \large\frac{\sin A+D_m/2}{\sin A/2} \\ (d)\;\mu= \sin \large\frac{\sin(A+D_m)}{\sin(A)} \end{array} $

1 Answer

The refractive index $\mu$ has been defined as the ratio of the speed of light in vacuum to the speed of light in the medium. It means that the refractive index of a given medium will be different for waves having wavelengths $3.8 \times 10^{-7}m$ and $5.8 \times 10^{-7}m$ because these waves travel with different speeds in the same medium. This variation of the refractive index of a material with wavelength is known as dispersion.

The angle between the emergent ray and the incident ray is known as the angle of deviation. 

We would now establish the relation between the angle of incident $i$, the angle of deviation $δ$ and the angle of prism $A$. Let us consider that a monochromatic beam of light $PQ$ is incident on  the face $AB$ of the principal section of the prism $ABC.$ On refraction, it goes along $QR$ inside the prism and emerges along $RS$ from face $AC$. Let $\angle{A} \equiv \angle{BAC}$ be the refracting angle of the prism. We draw normals $NQ$ and $MR$ on the faces $AB$ and $AC$, respectively and produce them backward to meet at $O$. Then you can easily convince yourself that $\angle{NQP} = \angle{i}, \angle{MRS} = \angle{e}, \angle{RQO} = \angle{r_1}$, and $\angle{QRO} = \angle{r_2}$ are the angle of incidence, the angle of emergence and the angle of refraction at the faces $AB$ and $AC$, respectively. The angle between the emergent ray $RS$ and the incident ray $PQ$ at $D$ is known as the angle of deviation ($δ$). 

Since $\angle{MDR} = \angle{δ}$, As it is the external angle of the triangle $QDR,$ we can write 
$\angle{δ} = \angle{DQR}+ \angle{DRQ} = (\angle{i} -\angle{r_1}) + (\angle{e}-\angle{r_2})$
or $\angle{δ} = (\angle{i}+\angle{e}) - (\angle{r_1} + \angle{r_2}) --- (1)$

You may recall that the sum of the internal angles of a quadrilateral is equal to $360^0$. In the quadrilateral $AQOR$, $\angle{AQO} = \angle{ARO} = 90^0$, since $NQ$ and $MR$ are normal on faces $AB$ and $AC$, respectively. 

Therefore $\angle{QAR}+\angle{QOR} = 180^0$ or $\angle{A}+ \angle{QOR} = 180^0 ---(2)$

But in $\Delta{QOR} \angle{OQR} + \angle{QRO} + \angle{QOR} = 180^0$
or $\angle{r_1} + \angle{r_2} + \angle{QOR} = 180^0 ---(3)$

On comparing Eqns. $(2)$ and $(3)$
we have $\angle{r_1} + \angle{r_2} = \angle{A} ---(4)$
Combining this result with Eqn. $(1)$, we have
$\angle{δ} = (\angle{i} + \angle{e}) - \angle{A}$
Or $\angle{i} + \angle{e} = \angle{A} + \angle{δ}---(5)$


If we vary the angle of incidence $i$, the angle of deviation $δ$ also changes it becomes minimum for a certain value of $i$ and again starts increasing as $i$ increases further. The minimum value of the angle of deviation is called angle of minimum deviation $(δm)$.

It depends on the material of the prism and the wavelength of light used. In fact,on angle of deviation may be obtained corresponding to two values of the angles of incidence. Using the principle of reversibility of light, we find the second value of angle of incidence corresponds to the angle of emergence $(e)$. In the minimum deviation position, there is only one value of the angle of incidence.
So we have $\angle{e} = \angle{i}$
Using this fact in Eqn. $(5)$ and replacing $δ$ by $δm$,

we have $\angle{i} = \angle{A} + \frac{\angle{δm}}{2} ---- (6)$
Applying the principle of reversibility of light rays and under the condition $\angle{e} = \angle{i}$, we can write
 $\angle{r_1} = \angle{r_2} = \angle{r}$, say
On substituting this result in Eqn. $(4)$, we get instruments.

$\angle{r} =$ $\frac{\angle{A}}{2}$ $ --- (7)$ 

The light beam inside the prism, under the condition of minimum deviation, passes symmetrically through the prism and is parallel to its base. The refractive index of the material of the prism is therefore given by 

$\mu = \frac{\sin i}{\sin r}  =$ $\frac{\sin(A + δm/2)}{\sin A/2}$ $---(8)$

The refractive index $\mu$ can be calculated using the Eqn.(8) for a monochromatic or a polychromatic beam of light. The value of $δm$ is different for different colours. It gives a unique value of the angle of incidence and the emergent beam is brightest for this incidence. For a prism of small angle $A$, keep $i$ and $r$ small, we can write

$\sin{i} = i, \sin r = r,$ and $\sin e = e$

Hence $\mu = \frac{\sin i}{\sin r_1} = \frac{i}{r_1}$ or $i = \mu r_1$
Also $\mu = \frac{\sin e}{\sin r_2} = \frac{e}{r_2}$ or $e = \mu r_2$

Therefore, $\angle{i} + \angle{e} =$ $\mu (\angle{r_1} + \angle{r_2})$   $(b)$
Using this result in Eqns. $(4)$ and $(5)$,
we get $\mu \angle{A} = \angle{A} + \angle{δ}$
or $\angle{δ} = (\mu - 1) \angle{A} ---(9)$

We know that $\mu$ depends on the wavelength of light. So deviation will also depend on the wavelength of light. That is why $δV$ is different from $δR$. Since the velocity of the red light is more than that of the violet light in glass, the deviation of the red light would be less as compared to that of the violet light. $δV > δR$. 

This implies that $\mu V > \mu R$. This change in the refractive index of the material with the wavelength of the light is responsible for dispersion phenomenon. 

Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.