What is Angle of Deviation in Reflection ?Explain With Diagram?

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Establish the Relation Among Deviation of a Ray, Angle of Incidence and Angle of Refraction

During reflection or refraction, the change in direction of light is called its deviation.

The angle between the refracted ray and the direction of the incident ray gives the measure of deviation.

Fig. shows the deviation of incident ray AO, which after refraction proceeds along OB instead of OC. So the deviation of ray, δ = ∠BOC.

We know, if the angle of incidence increases, the angle of refraction also increases.

i) For normal incidence, i = 0 thus r = 0 and so δ = 0 (minimum).

ii) For i = 90°, δ is maximum.
For refraction to a rarer medium from a denser medium, angle of refraction is greater than angle of incidence, i.e., r> i as shown in Fig.
δ = ∠BOC = ∠N’OB – ∠N’OC
= ∠N’OB – ∠AON = r – i

Refraction Of Light Through A Parallel Slab

ABCD is a glass slab whose faces AB and DC are parallel [Fig.]. A ray of light PQ in air is incident at Q on AB. After refraction the ray proceeds along QR and is incident at R on DC. Again being refracted the ray emerges in air along the path RS.

Let at Q the angle of incidence is and the angle of refraction is r₁ and at R the angle of incidence is r₂ and the angle of refraction is i₂.

Thus a ray incident on a parallel slab emerges parallel to itself in the same medium. Hence there is no deviation, only lateral displacement takes place.

Lateral displacement: In Fig., perpendicular RX drawn on PQ produced, gives the measure of the lateral displacement of PQ.

Let the thickness of the glass slab be t.

So the lateral displacement of a ray of light due to glass slab depends on

1. the thickness of the slab,
2. the angle of incidence and
3. the refractive Index of the material of the slab.

Special Case:

i) If angle of incidence i₁ is very small, then angle of refraction r₁ will be small too. In that case in equation (1) sin(i₁ – r₁) is replaced by (i₁ – r₁) and cosr₁ is replaced by l.
Thus lateral displacement,
RX = t(i₁ – r₁) = ti₁(1 – \(\frac{r_1}{i_1}\)) = ti₁(1 – \(\frac{1}{\mu}\))
(∵ i₁ and r₁ are very small, µ = \(\frac{\sin i_1}{\sin r_1}\) ≈ \(\frac{i_1}{r_1}\)]

ii) In case of normal incidence, i₁ = r₁ = 0. So, equation (1) provides RX = 0, i.e., for normal incidence of light on a parallel glass plate, no lateral displacement occurs.

iii) For grazing incidence on the surface AB, i₁ = 90°. So, equation (3) provides RX = t i.e., for angle of incidence 90° the value of lateral displacement equal to the thickness of the glass slab which is maximum.

Refraction of Light through Consecutive Parallel Media

Fig. shows three consecutive refractions in the parallel interfaces AB, CD and EF. These are the interfaces of media (a, b), (b, c) and (c, a) respectively.

For the refraction in AB, i₁ angle of incidence and r₁ = angle of refraction. For the second refraction in CD, r₁ = angle of incidence and r₂ = angle of refraction. For the third refraction
in EF, r₂ = angle of incidence and θ = angle of emergence..
Considering refraction at Q, R and S we can write,

Hence we conclude that if the first and final media are same then the incident ray and the emergent ray will be parallel to each other.
It means no lateral displacement occurs.

Numerical Examples

Example 1.
A ray of light is Incident on the plane face of a liquid at an angle 45° and due to refraction the ray is deviated through an angIe 15°. Calculate the refractive Index of the liquid.
Solution:
Angle of deviation of a ray for refraction from a rarer medium to a denser medium,
δ = i – r or, 15° = 45° – r or, r = 30°
∴ Refractive index of the liquid with respect to air,
µ = \(\frac{\sin i}{\sin r}\) = \(\frac{\sin 45^{\circ}}{\sin 30^{\circ}}\) = \(\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\) = \(\sqrt{2}\) = 1.414

Example 2.
The opposite faces of a glass slab of thickness 15 cm are parallel. If a ray of light is incident on the glass slab at an angle 60°, calculate the lateral displacement of the ray when it emerges from the slab. [µ of glass = 1.5]
Solution:
Lateral displacement of a ray of light incident at an angle i on a parallel glass slab having thickness t

Example 3.
A ray of light is refracted through a transparent sphere of refractive index µ in such a way that the ray passes through the ends of two radii Inclined at an angle θ with each other. If δ is the angle of deviation of the ray while passing through the sphere then prove that,
µ = \(\frac{\cos \frac{1}{2}(\theta-\delta)}{\cos \frac{\theta}{2}}\)
Solution:
Let the ray PQ incident at Q on the sphere, is refracted along QR and emerges from the sphere along RS
[Fig.]. Of the triangle OQR, OQ = OR (both are radii of the sphere).
∴ ∠OQR = ∠ORQ
Let angle of incidence of the ray at Q = ∠PQN = i and angle of refraction = ∠OQR = r

So the angle of incidence of the ray at R is r and angle of refraction i.e., angle of emergence is i.
Angle of deviation of the incident ray,
δ = ∠MTR = ∠TQR + ∠TRQ
= (i – r) + (i – r) = 2(i – r) …. (1)
From the triangle OQR we have, r + r + θ = 180°
or, 2r + θ = 180° or, r = 90° – \(\frac{\theta}{2}\) …. (2)
From equation (1) we get,

Example 4.
A parallel beam of rays of width 20 cm passing through a glass slab makes an angle ϕ = 60° with its plane surface. If the beam emerges into air from this surface then calculate the width of the emergent beam. Refractive Index of glass =1.8.
Solution:
The width of the beam of parallel rays = AB = 20 cm and the width of the beam of
emergent rays in air = CD.
From Fig., i = 30°

∴ _gµ_a = \(\frac{\sin 30^{\circ}}{\sin r}\)
or, sinr = \(\frac{1}{2} \cdot \frac{1}{g^{\prime} \mu_a}\) = \(\frac{a^\mu \mu_g}{2}\)
= \(\frac{1.8}{2}\) = 0.9
From the figure, ∠BAC = 30° and ∠ACD = r

Example 5.
Refractive index of water with respect to air is 1.33, refractive index of oil with respect to water is 1.45 and that of glass with respect to oil is 0.78. What is the refractive Index of glass with respect to air?
Solution: