Prism

A prism is a transparent optical object, typically made of glass or another transparent material, that bends (refracts) light as it passes through. It has flat, polished surfaces that are usually angled, and when light enters a prism, it is split into its constituent colors, creating a spectrum. This phenomenon is known as dispersion. Prisms are commonly used in science and technology, such as in optical devices, to study light behavior, split light into colors, or even bend light paths. They come in various shapes and sizes, with the triangular prism being the most familiar.

1.0Refraction Through Prism

An optical prism is a uniform, transparent solid material (like glass) with flat surfaces that bend light. When light passes through the prism, it encounters two flat, angled surfaces known as the 'refracting surfaces.' The angle formed between these two surfaces is referred to as the 'prism angle' or 'angle of the prism.

i : angle of incidence, e : angle of emergence.

$r_1$ and $r_2$ are refracting angles inside the prism.

A : apex angle of prism or  prism angle or Refracting angle of prism

Calculation of Prism Angle  

 For Ray-1 A=60°

 For Ray-2 A=70°

Important results regarding Prism

$A+(90°-r_1)+(90°-r_2)=180°$

$A=r_1+r_2$

2.0Deviation Produced By Prism

${\delta }_1=i-r_1\text{(Clockwise)}$

${\delta }_2=e-r_2\text{(Clockwise)}$

$\text{In both refraction, light ray deviated in same sense, hence,}$

${\delta }_{\text{net}}={\delta }_1+{\delta }_2\text{(Clockwise)}$

${\delta }_{\text{net}}=(i-r_1)+(e-r_2)$

${\delta }_{\text{net}}=(i+e)+(r_1+r_2)$

${\delta }_{\text{net}}=(i+e)-A\left(\because A=r_1+r_2\right)$

Example-1. Find out deviation produced by prism in a light ray?

Solution:   Snell’s law at first surface,

$1Sin60°=\sqrt{3}Sin{r}_1$

$\frac{\sqrt{3}}{2}=\sqrt{3}Sin{r}_1\Rightarrow r_1=30°$

$A=r_1+r_2$

$60°=30°+r_2\Rightarrow r_2=30°$

Snell’s law at second surface

 $\sqrt{3}Sin{r}_2=1Sine$

$\sqrt{3}Sin30°=Sine\Rightarrow e=60°$

${\delta }_{net}=i+e-A$

${\delta }_{net}=60°+60°-60°=60°$

Deviation Produced by Prism Graph

• The deviation caused by a prism is influenced by the angle of incidence (i), the angle of the prism (A), and the refractive index (μ)of the material.
•  $\delta$ is first decreasing and then increasing for i < e and i > e respectively.
• For a given deviation (excluding the minimum deviation), there are two possible values for the angle of incidence. If the angles of incidence and emergence are swapped, the same deviation is obtained, in accordance with the principle of reversibility of light.
• There is one and only one angle of incidence for which the angle of deviation is minimum (when i = e).
• Before i = e, δ  decreases more rapidly than it increases with i after i = e. The right hand side part of the graph is more tilted than the left hand side.

3.0Minimum Deviation in Prism

From graph condition of minimum deviation is i=e the, $r_1=r_2=\frac{A}{2}$

Relation Between Minimum Deviation And Refractive Index Of Prism

$\delta =i+e-A$

${\delta }_m=2i-A\Rightarrow i=\frac{{\delta }_m+A}{2}$

Snell’s law at first surface,

${\mu }_{s}Sini={\mu }_{p}Sin(\frac{A}{2})$

${\mu }_{s}Sin(\frac{{\delta }_m+A}{2})={\mu }_{p}Sin(\frac{A}{2})$

$\frac{{\mu }_p}{{\mu }_s}=\frac{Sin(\frac{{\delta }_m+A}{2})}{Sin(\frac{A}{2})}$

Note: For an isosceles / equilateral prism, light goes parallel to the base inside the prism.

Example: If refractive index of prism is $\sqrt{2}$ and refracting angle is 60°  then find out the minimum deviation produced by the prism.

Solution:

$\frac{{\mu }_p}{{\mu }_s}=\frac{Sin(\frac{{\delta }_m+A}{2})}{Sin(\frac{A}{2})}$$\Rightarrow \frac{\sqrt{2}}{1}=$$\frac{Sin(\frac{{\delta }_m+60^o}{2})}{Sin(\frac{60^o}{2})}$

$\sqrt{2}=\frac{\sin \left(\frac{{\delta }_m+60^{\circ }}{2}\right)}{\sin (30^{\circ })}\Rightarrow \frac{1}{\sqrt{2}}=\sin \left(\frac{{\delta }_m+60^{\circ }}{2}\right)\Rightarrow \frac{{\delta }_m+60^{\circ }}{2}=45^{\circ }\Rightarrow {\delta }_m=30^{\circ }$

4.0Maximum Deviation In A Prism

For ${\delta }_{max}$ either i=90° or e=90°

Then use ${\delta }_{net}$=i+e-A

5.0Condition of No Emergence

$(r_2{)}_{min}>{\theta }_c$

$A-r_{1max}>{\theta }_c$

When $i_{max}=90°$ then $r_{1max}={\theta }_c$

So, $A-{\theta }_c>{\theta }_c$

$A>2{\theta }_c$

6.0Condition of Always Emergence

$(r_2{)}_{max}<{\theta }_c$

$A-r_{1min}<{\theta }_c$

When $i_{min}=0°$ then  $r_{1min}=0°$

So, $A<{\theta }_c$

Conclusion

1. If $A\le {\theta }_c$  ; then light rays always cross the prism for any value of i.
2. If $A>2{\theta }_c$ ; then light ray never crosses the prism for any value of i.
3. If ${\theta }_c<A\le 2{\theta }_c$ ; then light rays may cross the prism or not depending on angle of incidence.

7.0Thin Prism

For a thin prism, prism angle (A) is very small.

In general A < 10°

In prism,

$\frac{{\mu }_p}{{\mu }_s}=\frac{Sin(\frac{{\delta }_m+A}{2})}{Sin(\frac{A}{2})}$

If prism is thin then A is very small & then we assume all angles are very small.

So, all apply $sin\theta \approx \theta$ then,

$\frac{{\mu }_p}{{\mu }_s}=\frac{Sin(\frac{{\delta }_m+A}{2})}{Sin(\frac{A}{2})}$$\Rightarrow \frac{{\mu }_p}{{\mu }_s}\times \frac{A}{2}=\frac{{\delta }_m+A}{2}$

${\delta }_m=(\frac{{\mu }_p}{{\mu }_s}-1)A$

The deviation for a small angled prism is independent of the angle of incidence.

8.0Sample Questions On Prism

Q-1. A ray of light incident on an equilateral glass prism shows minimum deviation of 30°,Calculate the speed of light through the prism.

Solution:

$\mu =\frac{\sin \left(\frac{{\delta }_m+A}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{{\mu }_p}{{\mu }_s}=\frac{\sin \left(\frac{30^{\circ }+60^{\circ }}{2}\right)}{\sin \left(\frac{60^{\circ }}{2}\right)}=\frac{\sin 45^{\circ }}{\sin 30^{\circ }}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}=1.414$

Velocity of light in glass

$v=\frac{c}{\mu }=\frac{3\times 10^8}{1.414}=2.18\times 10^8\text{m/s}$

Q-2. How does immersing a thin prism in water affect the angle of minimum deviation compared to when it is in air, and what is the resulting change in its value?. $a_{{\mu }_g}=\frac{3}{2}$$a_w=\frac{4}{3}$

Solution: Deviation produced by thin prism, $\delta =(\mu -1)A$

For thin prism placed in air,

${\delta }_1=\left(a_{{\mu }_g}-1\right)A=\left(\frac{3}{2}-1\right)A=\frac{1}{2}A$

For thin prism placed in water,

${\delta }_2=\left(w_{{\mu }_g}-1\right)A=\left(\frac{a_{{\mu }_g}}{a_{{\mu }_w}}-1\right)A=\left(\frac{\frac{3}{2}}{\frac{4}{3}}-1\right)A=\left(\frac{9}{8}-1\right)A=\frac{1}{8}A$

$\frac{{\delta }_2}{{\delta }_1}=\frac{1}{4}$

Q-3. An equilateral prism provides least deviation 46° in the air. Find out the refractive index of unknown liquid in which same prism gives least deviation 30°.

Solution:

$\frac{{\mu }_p}{{\mu }_s}=\frac{\sin \left(\frac{{\delta }_m+A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

For air, $\frac{{\mu }_p}{1}=\frac{\sin \left(\frac{46^{\circ }+60^{\circ }}{2}\right)}{\sin \left(\frac{60^{\circ }}{2}\right)}$...........(1)

For Liquid, $\frac{{\mu }_p}{{\mu }_l}=\frac{\sin \left(\frac{30^{\circ }+60^{\circ }}{2}\right)}{\sin \left(\frac{60^{\circ }}{2}\right)}$……(2)

From equations (1) and (2),

${\mu }_l=\frac{\sin 53^{\circ }}{\sin 45^{\circ }}=\frac{4\times \sqrt{2}}{5\times 1}=\frac{4\sqrt{2}}{5}$

Q-4. Find out the maximum value of the refractive index of a prism which permits the transmission of light through it when the refracting angle of the prism is 90°.

Solution: We can transmit the light through the prism when.

$A\le {\theta }_c\Rightarrow \frac{A}{2}\le {\theta }_c$

$\sin \left(\frac{A}{2}\right)\le \sin {\theta }_c\Rightarrow \sin \left(\frac{90^{\circ }}{2}\right)\le \frac{1}{\mu }$

$\mu \le \sqrt{2}\Rightarrow {\mu }_{\text{max}}=\sqrt{2}$

Q-5. A ray of light passing through a prism with a refractive index of $\sqrt{2}$ experiences minimum deviation. It is observed that the angle of incidence is twice the angle of refraction inside the prism. Determine the angle of the prism.

Solution:

For minimum deviation,

$r_1=r_2=\frac{A}{2}\text{and}i=2r_1\Rightarrow i=A$

Snell’s law at first surface, $1\times Sini={\mu }_{p}Sin{r}_1$

$\sin A=\sqrt{2}\sin \left(\frac{A}{2}\right)$

$2\sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right)=\sqrt{2}\sin \left(\frac{A}{2}\right)$

$\cos \left(\frac{A}{2}\right)=\frac{1}{\sqrt{2}}\Rightarrow \frac{A}{2}=45^{\circ }\Rightarrow A=90^{\circ }$