Refraction Through a Rectangular Glass Slab

Ray Optics is a branch of optics that describes light propagation in the form of rays. Ray Optics is also called geometrical optics which deals with the geometry of falling lights. this article is about refraction, its laws, refraction through glass slab and its various causes, and practice tracing the path of light rays passing through glass slab. Refraction is the property of the change in direction of the wave while entering from one medium to another medium. This change in direction is due to a change in the speed of the wave. 

What is Refraction? 

Refraction is the phenomenon of change in direction of a wave passing from one medium to another medium because of the change in its speed. Refraction refers to the bending of a wave when it is passed from one medium to another medium.

The wave bends because of the change in its speed from one medium to another medium. Here are two laws of Refraction that can be stated as:

• The incident ray refracted ray, and normal at a point of incidence lie on the same plane.
• The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant. This is known as Snell’s Law of refraction.

sin i / sin r = constant   

What is Glass Slab?

A glass slab is a substance with a cuboidal shape that is made of glass. A glass slab is made of glass with three dimensions length, breadth, and height.

Some important properties of a glass slab can be stated as:

• The glass slab does not deviate from the light rays passing through it.
• The glass slab does not disperse the light rays passing through it.
• Since it does not deviate and disperse the light rays passing through it that’s why the incident ray and the emergent ray emerging from the glass slab are parallel. The glass slab only produces a lateral displacement to the direction of light.

Refraction through the Glass Slab

• When the incident ray is incident on the surface of the glass slab from air to the glass making an angle of incidence i with the normal the refracted ray bends towards the normal as the ray enters from rarer to denser medium. 

Refraction through the Glass Slab

• Now, after traveling the glass slab the refracted ray makes an angle of incidence r on the other surface of the glass slab resulting in the emergent ray which bends away from the normal as it travels from glass (denser medium) to air (rarer medium) forming an angle of emergence e between the emergent ray and normal. 
• The emergent ray is parallel to the incident ray and the perpendicular distance between them is called lateral displacement. 
• Since the angle of incidence is equal to the angle of emergence, therefore the emergent ray is parallel to the incident ray. 
• In the glass slab, the light ray gets refracted two times firstly, from rarer to denser medium and secondly from denser to rarer medium. The displacement created in the emergent ray is due to refraction.

Therefore, the formula for the lateral displacement d is given by,

d = [t sin(i – r)]/ cos r = t sin(i – r) sec r 

where t is the thickness of the glass slab, i is the angle of incidence, r is the angle of refraction.

Experiment to Trace the path of a ray of light passing through a glass slab

Aim: Tracing the path of a ray of light passing through the glass slab.

Materials Required:- Drawing board, all pins, white paper, rectangular glass slab, protractor, scale, pencil, thumb pins.

Tracing the path of a ray of light  passing through a glass slab 

Procedure: 

1. Fix the white paper on the drawing board with the help of thumb pins.
2. Place the glass slab on the paper at the center and draw an outline of the glass slab and label it PQRS.
3. Draw a point E on PQ and draw perpendicular MN and label it as a normal ray.
4. Draw one angle of 30° with a protractor with MN. Fix pins A and B, 4-5 cm apart from each other on the ray that is obtained by formed angle.
5. Place the glass slab on the outline PQRS and see through the glass slab from side RS and fix C and D pins such that A, B, C, and D should lie on the same line.
6. Draw small circles on A, B, C, and D and remove the pins.
7. Join C, D such that it meets at point F on RS and draw perpendicular M’N’ at point F.
8. Now, join the points E and F.
9. Measure the angles formed at PQ and RS which are the angle of incidence, angle of refraction, and angle of emergence.
10. The lateral displacement d is obtained by extending the ray AB in the dotted line parallel to FCD and then, measuring it.

Conclusion:

• The angle of incidence is almost equal to the angle of emergence.
• As the light travels from rarer to a denser medium, the angle of refraction will be lesser than the angle of refraction.
• For different angles of incidence, the lateral displacement will be the same.
• The light bends towards the normal when it travels from rarer to denser medium.

 Different Cases of Refraction through Glass Slab

Case 1: Refraction when the object is in a denser medium and the observer is in a rarer medium

Consider a glass slab with refractive index and thickness μ and t respectively. The observer (eye) is in the air and the object (O) is in the glass slab.

Refraction when the object is in a denser medium and the observer is in a rarer medium

Hence for this case,

Virtual Depth = t / μ

and

Virtual displacement (OI) = OA – AI = t[1-(1/ μ)]

Case 2: Refraction through the successive slabs with different thicknesses and refractive index

Consider three successive slabs s₁, s₂, and s₃ with thickness t₁, t₂, t_3, and refractive indices μ₁, μ₂, and μ₃ respectively. These slabs are arranged one after the other with s₁ at the top, s₂ at the middle, and s₃ at the bottom. The object(O) is in s₃ and the observer (eye) is outside in the air (rarer medium).

Refraction through the successive slabs with different thicknesses and refractive index

Hence for this case,

Virtual depth (AI) = (t₁/ μ₁) + (t₂/ μ₂) + (t₃/ μ₃) + ……. 

and 

Virtual displacement (OI) = t₁[1 – (1/ μ₁)] + t₂[1 – (1/ μ₂)] + t₃[1 – (1/ μ₃)] + ………

Case 3: Refraction when object and observer both are in a rarer medium

Consider a glass slab with thickness t and refractive index μ. In this case, the object (O) and observer(eye) both are in a rarer medium (air) separated by the glass slab.

Refraction when object and observer both are in rarer medium

Hence for this case, 

Virtual displacement (OI) = [t – (1/ μ)]

Case 4: Refraction when the object is in a rarer medium and the observer is in a denser medium

Consider the observer(eye) is in water(denser medium) and the object (O) is in the air (rarer medium).

 Refraction when object is in a rarer medium and observer is in a denser medium 

Hence for this case, 

Real height / Virtual height = 1 / μ

or 

Virtual displacement (OI) = AI – AO = (μ – 1)AO 

Sample Questions

Question 1: Determine the position of the object kept at bottom of the glass slab with the thickness of 12 cm and the refractive index of glass is 1.5.

Answer:

The given case is refraction when the object is in a denser medium and the observer is in the rarer medium. 

We have to find virtual depth i.e. position of the object. 

Here, t = 12cm and μ = 1.5

Position of the object  = t / μ

= 12/1.5 

= 120/15

Position of the object  = 8 cm

Question 2: Two immiscible liquids of refractive indices 3/2 and 4/3 filled in the vessel up to 9 cm and 12 cm then, What will be the virtual depth of the bottom of the vessel.

Answer: 

The given case is like refraction through successive slab with different thickness and refractive index. 

Here, t₁ = 9cm, μ₁ = 3/2, t₂ = 12cm, and μ₂= 4/3

Virtual depth = (t₁/μ₁) + (t₂/μ₂)

= (9/(3/2)) + (12/(4/3))

= (18/3) + (36/4)

=  6 + 9

Virtual depth = 15 cm

Question 3: An object is kept at the bottom of the empty vessel and a microscope is focused to observe it. Now, water is filled in it up to 16cm then, at what amount microscope should move so that it can again focus on the object. The Refractive index is 4/3.

Answer: 

The given case is refraction when the object is in a denser medium and the observer is in the rarer medium. 

Here, t = 16 cm , μ = 4/3

The amount microscope should be moved so that it can again focus the object = Virtual displacement = t[1- (1/μ)] 

= 16[1 – {1/(4/3)}]

= 16 [1 – 3/4]

= 16 × (1/4)

The amount of microscope should be moved so that it can again focus the object = 4 cm.

Question 4: An air bubble in a glass slab (μ = 1.5) is 5 cm deep when viewed from one face and 2cm deep when viewed from the opposite face then find the thickness of the slab.

Answer: 

Given: 

μ = 1.5

Virtual depth v₁ when observed from one face = 5 cm

Virtual depth v₂ when observed from opposite face = 2 cm  

We have to find the real depth. This case is when the object is in a denser medium and the observer is in the rarer medium.

When observed from one face:  

v₁ = r₁/μ 

r₁ = v₁ × μ

= 5 × 1.5 = 7.5 cm

When observed from opposite face:  

v₂ = r₂/μ

r₂ = v₂ × μ

= 2 × 1.5 = 3.0 cm

Thickness of the glass slab = r₁ + r₂ = 7.5 + 3.0

= 10.5 cm

Question 5: What is lateral displacement. State the factors on which it depends.

Answer: 

The perpendicular distance between the incident ray and the resulting emergent ray is called the lateral displacement.

Factors on which lateral displacement depends:

• The thickness of the glass slab
• Angle of incidence
• Refractive index of glass 
• The wavelength of light used

Question 6: Why emergent ray is parallel to the incident ray in the refraction through the glass slab.

Answer: 

The emergent ray is parallel to the incident ray in the refraction through the glass slab because the angle of incidence and the angle of refraction are equal. 

Question 7: Find the lateral displacement when refraction through a glass slab takes place where the angle of incidence is 45° and angle of refraction is 30° and the thickness of the glass slab is 10 cm.

Answer: 

Here, t = 10cm, i = 45°, r = 30°

Lateral displacement (d) = [t sin(i – r)]/ cos r

= [10 sin( 45° – 30°)]/ cos 30°

= [10 sin 15°]/ cos 30°

= 10 × (sin 15° / cos 30°)

= 10 × (0.25/0.86)

= 10 × 0.29

Lateral displacement (d)  = 2.9 cm