WBCHSE Class 12 Physics Notes For Refraction Of Light Through A Parallel Slab

Refraction Of Light Through A Parallel Slab

A BCD is a glass slab whose faces AB ami DC are parallel. A ray of light 1>Q in the air is an Incident at Q ou AB. After refraction, the ruy proceeds along QR ami is Incident at R on DC. Again being refracted the ruy emerges In the air along the path RS.

Let at Q the angle of incidence is i₁ and the angle of refraction is r₁ and at R the angle of incidence is r₂ and the angle of refraction is i₂

For refraction at Q, \(a_g \mu_g=\frac{\sin i_1}{\sin r_1}\)

For refraction at R, \(g^{\mu_a}=\frac{\sin r_2}{\sin i_2}\)

Since, \(a^{\mu_g}=\frac{1}{g^{\mu_a}}\)

∴ \(\frac{\sin i_1}{\sin r_1}=\frac{\sin i_2}{\sin r_2}\)

According to r₁ = r₂ or, i₁= 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.

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Lateral displacement:

Perpendicular RX drawn on PQ produced gives the measure of the lateral displacement of PQ. Let the thickness of the glass slab be t

Now, sin ∠RQX =  \(\frac{R X}{Q R}\)

Or, RX =   \(Q R \sin \left(i_1-r_1\right)\)

[Since ∠RQX = i₁– r₁]

Now,  \(\cos r_1=\frac{Q T}{Q R}=\frac{t}{Q R}\)

Or, \(Q R=\frac{t}{\cos r_1}\)

RX = \(\frac{t}{\cos r_1} \cdot \sin \left(i_1-r_1\right)\)…………… (1)

= t\(\left[\frac{\sin i_1 \cos r_1-\cos i_1 \sin r_1}{\cos r_1}\right]\)

= t\(\left[\sin i_1-\frac{\cos i_1 \cdot \sin r_1}{\cos r_1}\right]\)…………… (2)

Now, if p is the refractive index of glass with respect to air then,

μ = \(\frac{\sin i_1}{\sin r_1}\)

Or, \(\sin r_1=\frac{\sin i_1}{\mu}\)

∴ \(\cos r_1=\sqrt{1-\sin ^2 r_1}\)

= \(\sqrt{1-\frac{\sin ^2 i_1}{\mu^2}}=\frac{\sqrt{\mu^2-\sin ^2 i_1}}{\mu}\)

∴  RX = t\(\left[\sin i_1-\frac{\cos i_1 \sin i_1 \mu}{\mu \sqrt{\mu^2-\sin ^2 i_1}}\right]\)

= \(\sin i_1\left[1-\frac{\cos i_1}{\sqrt{\mu^2-\sin ^2 i_1}}\right]\)…………(3)

So the lateral displacement of a ray of light due to a 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:

If the angle of incidence, i₁ is very small, then the angle of refraction r, will be small too. In that case In equation (1) sin(i₁-r₁) is replaced by (,-r,) and cosr₁, is replaced by 1 .

Thus the lateral displacement

RX = t\(\left(i_1-r_1\right)=t i_1\left(1-\frac{r_1}{i_1}\right)=t i_1\left(1-\frac{1}{\mu}\right)\)

Since, [i₁ and r₁ are very small , \(\mu=\frac{\sin i_1}{\sin r_1} \approx \frac{i_1}{r_1}\)]

In the 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.

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

Refraction of Light through Consecutive Parallel Media

Three consecutive refractions in the parallel interfaces AB, CD and EF. These are the interfaces of media (o, 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

_aμ_b =  \(a^{\mu_b}=\frac{\sin i_1}{\sin r_1}\)

_bμ_c =  \(b^{\mu_c}=\frac{\sin r_1}{\sin r_2}\) and,

_cμ_a =  \(c_a=\frac{\sin r_2}{\sin \theta}\)

So, \({ }_a \mu_b \times{ }_b \mu_c \times{ }_c \mu_a=\frac{\sin i_1}{\sin r_1} \times \frac{\sin r_1}{\sin r_2} \times \frac{\sin r_2}{\sin \theta}\)

i.e, \(\frac{\mu_b}{\mu_a} \times \frac{\mu_c}{\mu_b} \times \frac{\mu_a}{\mu_c}=\frac{\sin i_1}{\sin \theta} \quad \text { or, } \frac{\sin i_1}{\sin \theta}\) = 1 Or, θ = i₁

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

Refraction Of Light Refraction Of Light Through A Parallel Slab Numerical Examples

Example 1. A ray of light Is incident on the plane face of liquid at an angle of 45° and due to refraction the.ray Deviates through an angle of 15°. Calculate the refractive Index of the liquid.
Solution:

The angle of deviation of a ray for refraction from a medium to a denser medium,

δ = i-r or, 15° = 45°- r or, r = 30°

The refractive index of the liquid with respect to air,

⇒ \(\mu=\frac{\sin i}{\sin r}\)

⇒ \(\frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\)

⇒ \(\sqrt{2}\)

= 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 of 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

= \(t \sin i\left[1-\frac{\cos i}{\sqrt{\mu^2-\sin ^2 i}}\right]\)

[ μ= 1.5]

= \(15 \sin 60^{\circ}\left[1-\frac{\cos 60^{\circ}}{\sqrt{\left(\frac{3}{2}\right)^2-\sin ^2 60^{\circ}}}\right]\)

= \(=15 \times \frac{\sqrt{3}}{2}\left[1-\frac{\frac{1}{2}}{\sqrt{\frac{9}{4}-\frac{3}{4}}}\right]\)

= 7.69 cm

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 of θ with each other. If  δ is the angle of deviation of the ray while passing through the sphere then prove that \(\mu=\frac{\cos \frac{1}{2}(\theta-\delta)}{\cos \frac{\theta}{2}}\)
Solution:

Let’ the ray PQ incident at Q,!oh the sphere is refracted along QR and emerges from the sphere along RS

Of the triangle OQR, OQ = OR (both are radii of the

∴ ∠OQR = ∠ORQ

Let the angle of incidence of

The ray at Q = ∠PQN = i and angle of refraction

= ∠OQR = r

So the angle of incidence of p the ray at R is r and the angle of refraction i.e., angle of emergence is i.

The 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,

⇒ \(\frac{\delta}{2}\) = i-r

Or, i= \(\frac{\delta}{2}+r\) = \(\frac{\delta}{2}+90^{\circ}-\frac{\theta}{2}\)

= 90 + \(\frac{\delta}{2}-\frac{\theta}{2}\)

Considering refraction at Q,

μ = \(\frac{\sin i}{\sin r}=\frac{\sin \left(90^{\circ}+\frac{\delta}{2}-\frac{\theta}{2}\right)}{\sin \left(90^{\circ}-\frac{\theta}{2}\right)}\)

= \(\frac{\sin \left\{90^{\circ}-\frac{1}{2}(\theta-\delta)\right\}}{\cos \frac{\theta}{2}}\)

= \(\frac{\cos \frac{1}{2}(\theta-\delta)}{\cos \frac{\theta}{2}}\)

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 the air from this surface then calculate the width of he emergent beam. The 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.

⇒ \({ }_g \mu_a=\frac{\sin 30^{\circ}}{\sin r}\)

Or, \(\frac{1}{2} \cdot \frac{1}{g^\mu}\)

= \(\frac{a^\mu{ }^ g}{2}\)

= \(\frac{1.8}{2}\) = 0.9

From the ∠BAC =  30° and ∠AC.P = r

Now, Cos 30° = \(\frac{A B}{A C}\)

Cos r, \(\frac{C D}{A C}\)

⇒ \(\frac{\cos 30^{\circ}}{\cos r}=\frac{A B}{C D}\)

Or, \(\frac{A B \cos r}{\cos 30^{\circ}}\)

= \(\frac{20 \sqrt{1-\sin ^2 r}}{\frac{\sqrt{3}}{2}}\)

= \(\frac{40}{\sqrt{3}} \sqrt{1-0.81}\) = \(\frac{40}{\sqrt{3}} \times \sqrt{0.19}\)

= 10.064 cm

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

Here,  _aμ_w = 1.33; _wμ_o = 1.45; _dμ_g = 0.78

We know,

_aμ_w×_wμ_o×_dμ_g×_gμ_a = 1

Or, _aμ_w×_wμ_o×_dμ_g×_gμ_a= \(\frac{1}{g^{\prime \prime} a}=a_a \mu_g\)

Or, 1.33 × 1.45 × 0.78

Or,_aμ_g = 1.5