Refraction Of Light

Refraction through a glass slab

Definition: Refraction is the bending of light when it travels from one medium to another medium.
Refractive Index (n): Ratio of the speed of light in vacuum to the speed of light in a medium.
Snell’s Law: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the refractive index.
Formula for Snell’s Law: n₁ sin(θ₁) = n₂ sin(θ₂)
Total Internal Reflection: When light is incident from a denser medium to a rarer medium at an angle greater than the critical angle, it is totally reflected back into the denser medium.
Critical Angle (θc): The angle of incidence for which the angle of refraction is 90 degrees.

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Refraction through a glass slab

Refraction through a glass slab can be understood using the principles of refraction.
When light passes through a glass slab, it undergoes refraction twice – once at the entry surface and once at the exit surface of the slab.
The incident ray, refracted ray, and the normal at both the surfaces lie in the same plane.
The deviation produced by a glass slab depends on the refractive index of the glass and the angle of incidence.
The lateral shift (displacement) produced by a glass slab can be calculated using the formula: D = t (μ - 1) α
- D: Lateral shift or displacement
- t: Thickness of the glass slab
- μ: Refractive index of the glass
- α: Angle of incidence

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Refraction through a glass slab

The lateral shift is directly proportional to the thickness of the slab, angle of incidence, and refractive index.
If the angle of incidence is zero, there will be no lateral shift.
The lateral shift is maximum when the angle of incidence is 90 degrees.
The lateral shift increases with an increase in the thickness of the glass slab.
The direction of the lateral shift depends on the orientation of the glass slab with respect to the incident ray.

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Refraction through a glass slab

Example: Light is incident normally on a glass slab (θ₁ = 0 degrees).
The lateral shift is given by D = t (μ - 1) α.
As the angle of incidence is zero, the lateral shift is also zero.
Example: Light is incident at an angle of 60 degrees on a glass slab of refractive index 1.5.
The lateral shift can be calculated using the formula D = t (μ - 1) α.

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Refraction through a glass slab

The refractive index of a medium depends on the wavelength of light.
The refractive index of a medium is different for different colors of light, leading to dispersion.
Dispersion: The phenomenon of separating white light into its constituent colors.
The refractive index of a medium is highest for violet light and lowest for red light.
The angle of refraction is different for different colors when passing through a prism, leading to the formation of a spectrum.

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Refraction through a glass slab

The lateral shift produced by a glass slab is independent of the color of light passing through it.
However, the angles of incidence and refraction vary for different colors, leading to dispersion.
The phenomenon of dispersion is utilized in the formation of a spectrum in various optical devices.
Examples of optical devices utilizing dispersion: Prism, spectrometer, rainbow formation, etc.

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Refraction through a glass slab

Application: Glass slabs are used in optical devices to redirect and control the path of light.
Prism: A triangular glass slab that refracts light and produces a spectrum.
Spectrometer: An instrument that uses a prism to separate and analyze the colors of light.
Optical fibers: Long, thin strands of glass that transmit light through total internal reflection.
Thick lenses: Used in spectacles and camera lenses to correct vision problems and focus light.
These applications rely on the properties of refraction and the behavior of light passing through glass slabs.

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Refraction through a glass slab

Summary:
- Refraction is the bending of light when it travels from one medium to another.
- Snell’s Law relates the angles of incidence and refraction to the refractive index.
- Refraction through a glass slab occurs twice, leading to a lateral shift.
- The lateral shift depends on the thickness, angle of incidence, and refractive index.
- Dispersion is the separation of white light into its constituent colors due to different refractive indices.
- Optical devices like prism and spectrometer use refraction and dispersion for various applications.

Slide 11

Recap:
- Refraction is the bending of light when it travels from one medium to another.
- Snell’s Law relates the angles of incidence and refraction to the refractive index.
- Refraction through a glass slab occurs twice, leading to a lateral shift.
- The lateral shift depends on the thickness, angle of incidence, and refractive index.
- Dispersion is the separation of white light into its constituent colors due to different refractive indices.

Slide 12

Example 1: Light enters a glass slab (μ = 1.5) with an angle of incidence of 30 degrees. Calculate the angle of refraction.
Solution: Using Snell’s Law, n₁ sin(θ₁) = n₂ sin(θ₂), we have sin(θ₂) = (n₁ / n₂) * sin(θ₁)
Plugging in the values, sin(θ₂) = (1 / 1.5) * sin(30°) = 0.577 * 0.5 = 0.2888
Taking the inverse sine of 0.2888, we get θ₂ ≈ 17.54°.

Slide 13

Example 2: A light ray passes through a glass slab (μ = 1.6) at an angle of incidence of 45 degrees. Calculate the angle of refraction.
Solution: Using Snell’s Law, n₁ sin(θ₁) = n₂ sin(θ₂), we have sin(θ₂) = (n₁ / n₂) * sin(θ₁)
Plugging in the values, sin(θ₂) = (1 / 1.6) * sin(45°) = 0.625 * 0.7071 = 0.4418
Taking the inverse sine of 0.4418, we get θ₂ ≈ 26.8°.

Slide 14

Example 3: A light ray enters a glass slab (μ = 1.8) at an angle of incidence of 60 degrees. Calculate the angle of refraction.
Solution: Using Snell’s Law, n₁ sin(θ₁) = n₂ sin(θ₂), we have sin(θ₂) = (n₁ / n₂) * sin(θ₁)
Plugging in the values, sin(θ₂) = (1 / 1.8) * sin(60°) = 0.556 * 0.866 = 0.481
Taking the inverse sine of 0.481, we get θ₂ ≈ 29.1°.

Slide 15

Example 4: A light ray enters a glass slab (μ = 1.5) parallel to the base of the slab. Calculate the angle of refraction.
Solution: When the light ray enters parallel to the base, the angle of incidence is 90 degrees. The angle of refraction can be calculated using Snell’s Law, sin(θ₂) = (n₁ / n₂) * sin(θ₁)
Plugging in the values, sin(θ₂) = (1 / 1.5) * sin(90°) = 0.667 * 1 = 0.667
Taking the inverse sine of 0.667, we get θ₂ ≈ 41.81°.

Slide 16

Recap:
- Refraction can be described using Snell’s Law: n₁ sin(θ₁) = n₂ sin(θ₂)
- Examples were provided to calculate the angle of refraction in different scenarios.
- The angle of refraction depends on the refractive indices and the angle of incidence.
- When the angle of incidence is 90 degrees, the angle of refraction can be calculated as sin(θ₂) = (n₁ / n₂).

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Refraction through a glass slab

Example 5: A glass slab (μ = 1.5) of thickness 2 cm is placed in air (μ = 1). A light ray enters the slab at an angle of incidence of 45 degrees. Calculate the lateral shift.
Solution: The lateral shift can be calculated using the formula D = t (μ - 1) α.
Plugging in the values, D = 0.02 m (1.5 - 1) 45° ≈ 0.3 m.

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Refraction through a glass slab

Example 6: A glass slab (μ = 1.6) is immersed in a medium with a refractive index of 1.3. A light ray enters the slab at an angle of incidence of 30 degrees. Calculate the angle of refraction in the medium.
Solution: Using Snell’s Law, n₁ sin(θ₁) = n₂ sin(θ₂), we have sin(θ₂) = (n₁ / n₂) * sin(θ₁).
Plugging in the values, sin(θ₂) = (1.6 / 1.3) * sin(30°) ≈ 1.231 * 0.5 = 0.616.
Taking the inverse sine of 0.616, we get θ₂ ≈ 38.04°.

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Refraction through a glass slab

Example 7: A light ray passes through a glass slab (μ = 1.4) at an angle of incidence of 60 degrees and undergoes a lateral shift of 0.08 m. Calculate the thickness of the slab.
Solution: Rearranging the lateral shift formula, we have t = D / ((μ - 1) α).
Plugging in the values, t = 0.08 m / ((1.4 - 1) 60°) ≈ 1.43 m.

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Refraction through a glass slab

Example 8: A glass slab is immersed in water (μ = 1.33). A light ray enters the slab at an angle of incidence of 45 degrees. Calculate the angle of refraction in water.
Solution: Using Snell’s Law, n₁ sin(θ₁) = n₂ sin(θ₂), we have sin(θ₂) = (n₁ / n₂) * sin(θ₁).
Plugging in the values, sin(θ₂) = (1.5 / 1.33) * sin(45°) ≈ 1.128 * 0.7071 = 0.797.
Taking the inverse sine of 0.797, we get θ₂ ≈ 53.5°.

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Refraction through a glass slab

Example 9: A light ray passes through a glass slab. The angle of incidence is twice the angle of refraction. Calculate the refractive index of the medium.
Solution: Using Snell’s Law, n₁ sin(θ₁) = n₂ sin(θ₂), we can write the given condition as θ₁ = 2θ₂.
Substituting the equation into Snell’s Law, n₁ sin(2θ₂) = n₂ sin(θ₂).
Rearranging the equation, n₂ / n₁ = sin(2θ₂) / sin(θ₂) = 2 cos(θ₂).
As θ₂ is the angle of incidence and θ₁ is the angle of refraction, n₂ / n₁ = 2 cos(θ₁).
Therefore, the refractive index of the medium is 2 cos(θ₁).

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Refraction through a glass slab

Example 10: A glass slab (μ = 1.5) is placed in a medium with a refractive index of 1.33. Calculate the angle of incidence at which the light ray will emerge parallel to the base of the slab.
Solution: When the light ray emerges parallel to the base, the angle of refraction is 90 degrees. The angle of incidence can be calculated using Snell’s Law, n₁ sin(θ₁) = n₂ sin(θ₂).
Plugging in the values, sin(90°) = (1.5 / 1.33) * sin(θ₁).
Simplifying the equation, 1 = (1.5 / 1.33) * sin(θ₁) ≈ 1.128 * sin(θ₁).
Taking the inverse sine of 1 / 1.128, we get θ₁ ≈ 87°.

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Refraction through a glass slab

Summary:
- Examples were provided to calculate the lateral shift, angle of refraction, thickness of the slab, and refractive index in various scenarios.
- Equations such as D = t (μ - 1) α, n₁ sin(θ₁) = n₂ sin(θ₂), and n₂ / n₁ = 2 cos(θ₁) were used in solving the problems.
- Understanding the behavior of light passing through a glass slab is essential for various optical applications.