Refraction of Light Through a Prism

1.0Prism

An optical prism is a homogeneous solid transparent medium (such as glass) with flat surfaces that refract light. In prism light ray courses two plane inclined surfaces. These surfaces are called the ‘refracting surfaces’ and the angle between them is called the ‘refracting angle’ or the ‘angle of prism’.

i : angle of incidence
e : angle of emergence
r₁ & r₂ are refracting angle inside the prism.
A : apex angle OR prism angle
OR Refracting angle of prism.

2.03-D view

Refraction through a prism:
PQ and PR are refracting surfaces.
∠QPR = A is called refracting angle or the angle of prism (also called Apex angle).
δ = angle of deviation

Angle of Deviation (δ)

It is the angle between the emergent and the incident ray. In other words, it is the angle through which the incident ray turns while passing through a prism.

δ = (r₁ + r₂) − A
= (i − r₁) + (e − r₂)
= i + e − A

Condition of No Emergence

The light will not emerge out of a prism for all values of angle of incidence i at face AB for
i = i_max = 90°, at face AC r₂ > θc

From Snell’s law at AB,

1 × sin 90° = μ sin r₁
⇒ sin r₁ = (1/μ)

⇒ r₁ = θc

r₁ + r₂ > 2θc 

But r₁ + r₂ = A

As A ≥ 2θc

or,  sin(A/2) ≥ sin θc

μ ≥ 1/sin(A/2) or μ ≥ cosec(A/2)

i.e., a ray of light will not emerge out of a prism (whatever be the angle of incidence) if

A ≥ 2θc
i.e., if μ ≥ cosec(A/2)

3.0Critical Angle

It is the angle of prism above which incident light ray on first surface will not emerge out from the second surface for all possible values of angle of incidence
A = 2θ_c

4.0Condition of Grazing Emergence

If a ray can emerge out of a prism, the value of angle of incidence i for which angle of emergence e = 90° is called condition of grazing emergence.                            

Now from Snell's law at AB,

Note:

sin i = µ sin r₁
sin i = µ sin (A – θ_c)
= μ [sin A cos θ_c - cos A sin θ_c]
= µ [sin A√(1-sin² θ_c) - cos A sin θ_c]
= [sin A √(1 - 1/μ²) - cos A/μ] = [√(μ² - 1) sin A - cos A]/μ

i = sin⁻¹ [√(μ² −1) sin A - cos A]

Note:

The light will emerge out of a given prism only if the angle of incidence is greater than the condition of grazing emergence.

5.0Condition of maximum deviation

Deviation will be maximum when
i_max = 90°

So, δ_max = i_max + e - A
= 90° + e - A

However, when i = 90°,
For "AB"
1× sin 90° = µ sin r₁

sin r₁ = 1/μ
r₁ = θc

Again, r₁ + r₂ = A
⇒ r₂ = A - θc

Now for 'AC'
µ sin r₂ = 1 sin e
sin e = μ sin (A – θc)
e = sin⁻¹ [µ sin (A – θc)]

Thus, δ_max = 90° + sin⁻¹[µ sin(A – θ_c)] – A

Note:
This situation is reverse of grazing emergence and may also be viewed as deviation at grazing incidence.

As δ = i + e - A
δ_min = 2i - A

Using Snell's law:
1 x sin i = µ sin r_₁
and µ sin r_₂ = 1 x sin i

sin r_₁ = sin r_₂
r_₁ = r_₂ = r

δ_min = (2i - A)
where r = A/2

$\mu =sini/sinr=sin({\delta }_{m}in+A)/2/sin(A/2)$

Note:
In the condition of minimum deviation the light ray passes through the prism symmetrically, i.e., the light ray in the prism becomes parallel to its base.

Proof: We know, $\delta +A=i+e$

And $r_1+r_2=A$

Again $\sin i=\mu \sin r_1$

And $\sin e=\mu \sin r_2$

for minimum deviation, $\frac{d\delta }{di}=0$

$\therefore \frac{d\delta }{di}=\frac{di}{di}+\frac{de}{di}=1+\frac{de}{di}=0$

I.e. $\frac{de}{di}=-1$

Also $d{r}_1=-d{r}_2$

$\cos i,di=\mu \cos r_1,d{r}_1$

and

$\cos e,de=\mu \cos r_2,d{r}_2$

$\therefore \frac{\cos i,di}{\cos e,de}=\frac{\cos r_1,d{r}_1}{\cos r_2,d{r}_2}$

$\Rightarrow \frac{{\cos }^{2}i}{{\cos }^{2}e}$

$\frac{{\cos }^2{r}_1}{{\cos }^2{r}_2}$

$\frac{1-{\sin }^2{r}_1}{1-{\sin }^2{r}_2}$

$\frac{1-\frac{1}{{\mu }^2}{\sin }^{2}i}{1-\frac{1}{{\mu }^2}{\sin }^{2}e}$

$\frac{1-{\sin }^{2}i}{1-{\sin }^{2}e}$

$\frac{{\mu }^2-{\sin }^{2}i}{{\mu }^2-{\sin }^{2}e}$

$\Rightarrow {\sin }^{2}i={\sin }^{2}e$

$\therefore i=e$

6.0Graphical Representation of Angle of Deviation

The deviation produced by a prism depends on angle of incidence i, angle of prism A, refractive index of material μ.

(i) δ is first decreasing and then increasing for i<e and i > e respectively.

(ii) For one δ (except δ_min) there are two values of angle of incidence. If i and e are interchanged then we get the same value of δ because of reversibility principle of light

(iii) Before i = e, δ decreases more rapidly than it increases with i after i = e.

7.0Thin Prisms

In thin prism the distance between the refracting surfaces is negligible and the angle of prism (A) is very small.

Since A = r_₁ + r_₂, therefore, r_₁ and r_₂ both are small and the same is true for i_₁ and i_₂

According to Snell's law:
sin i = µ sin r_₁
⇒ i = μr_₁

sin e = µ sin r_₂
⇒ e = μr_₂

Note:
Deviation, δ = (i_₁-r_₁) + (e-r_₂) = (r_₁ + r_₂) (μ – 1) = A (μ – 1)

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

If object is real

1. Image is virtual
2. Shifting of image is OI

8.0Dispersion of Light

The angular splitting of a ray of white light into a number of components and spreading in different directions is called Dispersion of Light. [It is for whole Electro Magnetic Wave in totality]. This phenomenon is because waves of different wavelength move with same speed in vacuum but with different speeds in a medium.

Therefore, the refractive index of a medium depends slightly on wavelength also. This variation of refractive index with wavelength is given by Cauchy's formula.

Cauchy's formula: μ (λ) = a + b/λ² where a and b are positive constants of a medium.

Note:
Such phenomenon is not exhibited by sound waves.

Angle between the rays of the extreme colours in the refracted (dispersed) light is called angle of dispersion.
θ = δᵥ - δᵣ (Fig. (a))

Fig (a) and (c) represents dispersion, whereas in fig. (b) there is no dispersion.

For prism of small 'A' and with small 'i':
θ = δᵥ – δᵣ = (μᵥ – μᵣ) A

Deviation of beam (also called mean deviation)
δ = δᵧ = (μ – 1)A
μᵥ, μᵣ and μᵧ are R. I. of material for violet, red and yellow colours respectively.

Note:

Numerical data reveals that if the average value of μ is small μᵥ – μᵣ is also small and if the average value of μ is large μᵥ - μᵣ is also large. Thus, larger the mean deviation, larger will be the angular dispersion.

9.0Dispersive power (ω):

Dispersive power (ω) of the medium of the material of prism is given by: $\omega =\frac{{\mu }_v-{\mu }_r}{{\mu }_y-1}$

ω is the property of a medium.

For small angled prism (A ≤ 10°) with light incident at small angle i :

$\frac{{\mu }_v-{\mu }_r}{{\mu }_y-1}=\frac{{\delta }_v-{\delta }_r}{{\delta }_y}=\frac{\theta }{{\delta }_y}=\frac{\text{Angular dispersion}}{\text{Deviation of mean ray (yellow)}}$

$\left[{\mu }_y=\frac{{\mu }_v+{\mu }_r}{2}\text{if }{\mu }_y\text{ is not given in the problem}\right]$

• μ − 1 = Refractive index of the medium for the corresponding colour.

Dispersion without deviation (Direct Vision Combination)

The condition for direct vision combination is:

${\mu }_v-1]A=[{\mu }_v^{\prime }-1]A^{\prime }=⟺=\left[\frac{{\mu }_v={\mu }_r}{2}-1\right]A=\left[\frac{{\mu }_v^{\prime }+{\mu }_r^{\prime }}{2}-1\right]A^{\prime }$

Two or more prisms can be combined in various ways to get different combination of angular dispersion and deviation.

10.0Deviation without dispersion (Achromatic Combination)

Condition for achromatic combination is: (μᵥ – μᵣ) A = (μ'ᵥ – μ'ᵣ) A'

or ωA = ω'A'

Solved Examples

Problem 1:

A ray of light is incident at an angle of 60° on one face of a prism which has an angle of 30°. The ray emerging out of the prism makes an angle of 30° with the incident ray. Show that the emergent ray is perpendicular to the face through which it emerges and calculate the refractive index of the material of the prism.

Solution:

According to given problem A=30°, i_₁ = 60°
and δ = 30° and as in a prism

δ = (i_₁+i_₂)-A, 30° = (60+i_₂) -30 i.e., i_₂ = 0

So the emergent ray is perpendicular to the face from which it emerges

Now as i_₂ = 0, r_₂ = 0

But as r_₁ + r_₂ = A,
r₁ = A = 30°

So at first face
1 x sin 60° = µ sin 30° i.e., µ = √3

Problem 2:

A right prism is to be made by selecting a proper material and angle A and B (≤ A) as shown in Fig. It is desired that a ray of light incident on the face AB emerges parallel to the incident direction after two internal reflections.

(a) What should be the minimum refractive index μ for this to be possible
(b) For µ = (5/3) is it possible to achieve this with angle B = 30°.

Solution:

(a) As ∠C = 90°, ∠A + ∠B = 90° with B ≤ A. Now TIR at P will take place if iₐ > ${\theta }_c$ and TIR at Q will take place if iᵦ > ${\theta }_c$

So that $(i_A+i_B>2{\theta }_c)$

But from geometry of Fig $(i_A=A)$ and $(i_B=B)$

So,

$(A+B>2{\theta }_c)$

or

$(90^{\circ }>2{\theta }_c)[as(A+B=90^{\circ })]$

or

$(45^{\circ }>{\theta }_c)$

i.e.,

$(\sin 45^{\circ }>\sin {\theta }_c)$

or

$\frac{1}{\sqrt{2}}>\frac{1}{\mu }$

$\text{or}$

$\mu >\sqrt{2}$
i.e., $({\mu }_{\min }=\sqrt{2})$

(b)

For $(B=30^{\circ })$ TIR at Q will take place only if $(30^{\circ }>{\theta }_c)$

Or $(\sin 30^{\circ }>\sin {\theta }_c)$

Or $\frac{1}{2}>\frac{1}{\mu }$

$\Rightarrow \mu >2$
$\Rightarrow {\mu }_{\min }=2$

But as here ( $\mu =(5/3)<2$ ). So, TIR at Q is not possible with ( $\mu =(5/3)$ ) and ( $B=30^{\circ }$ ).

Problem 3:

Refractive index of glass for red and violet colours are 1.50 and 1.60 respectively. Find :

(a) the refractive index for yellow colour, approximately
(b) dispersive power of the medium.

Solution:

(a)

${\mu }_y\approx \frac{{\mu }_V+{\mu }_R}{2}=\frac{1.50+1.60}{2}=1.55$

(b)

$\omega =\frac{{\mu }_V-{\mu }_R}{{\mu }_y-1}=\frac{1.60-1.50}{1.55-1}=0.18$