Refraction of Light Waves Notes

Refraction of Light Waves

Refraction is the bending of a light ray as it passes from one medium to another with different densities, causing a change in direction. This bending is due to the difference in the speed of light in the two media.

Terms Associated with Refraction

• Incident ray: The path of light in the first medium (e.g., air).
• Refracted ray: The path of light in the second medium (e.g., glass).
• Emergent ray: The path of light as it emerges from the second medium.
• Angle of incidence (i): The angle between the incident ray and the normal to the surface at the point of incidence (O).
• Angle of refraction (r): The angle between the refracted ray and the normal to the surface at the point of incidence (O).
• Emergent angle (e): The angle between the emergent ray and the normal.
• Angle of deviation (d): The angle between the original direction of the incident ray and the direction of the refracted ray, $d = i - r$.

Note: When a ray travels from a less dense to a denser medium, it bends towards the normal. Conversely, when it travels from a denser to a less dense medium, it bends away from the normal.

Laws of Refraction

1. The incident ray, the refracted ray, and the normal at the point of incidence all lie in the same plane.
2. Snell's Law: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media. This constant is the refractive index ($\mu$) of the second medium with respect to the first.
   $\mu = \frac{\sin i}{\sin r}$

Real Depth, Apparent Depth, and Refractive Index

The refractive index ($\mu$) can also be expressed as the ratio of real depth to apparent depth:

$\mu = \frac{\text{real depth}}{\text{apparent depth}}$

Apparent displacement (or lateral displacement) is the difference between the real depth and the apparent depth:

$\text{Apparent displacement} = \text{real depth} - \text{apparent depth}$

Applications of Refraction

1. Apparent depth of a swimming pool: A swimming pool appears shallower than its actual depth due to refraction.
2. Apparent bending of a stick partially immersed in water: The immersed part of the stick appears broken and bent upwards because rays from different points on the object below the water surface are bent away from the normal when emerging from water (denser) to air (less dense).
3. Bringing objects into view: A coin in a bowl not visible to an observer can be brought into view by pouring water into the bowl. The water causes refraction, allowing the observer to see the coin.

Refractive Index Details

• When light travels from air to glass, the refractive index is given as:

  $<em>{a}\mu</em>{g} = \frac{\sin i}{\sin r}$.

• While speed, wavelength, and direction of light change from one medium to another, the frequency remains the same.

  $<em>{a}\mu</em>{g} = \frac{v<em>{1}}{v</em>{2}} = \frac{\lambda<em>{1}}{\lambda</em>{2}}$

• For light traveling from glass to air:

  $<em>{g}\mu</em>{a} = \frac{\sin r}{\sin i}$

  $<em>{a}\mu</em>{g} = \frac{1}{<em>{g}\mu</em>{a}}$

• The refractive index is a measure of how light propagates through a material. It indicates the extent to which a medium allows light to pass through it. A higher refractive index means light travels slower, causing a greater change in direction within the material.

• Refractive index of glass is approximately 1.5, and for water, it is about 1.33.

Total Internal Reflection and Critical Angle

The critical angle is the angle of incidence at which the angle of refraction is 90 degrees when light travels from a denser to a less dense medium.

Total Internal Reflection (TIR) occurs when the critical angle is exceeded for light traveling from a denser to a less dense medium. When the angle of incidence is greater than the critical angle, the light is completely reflected back into the denser medium.

Relationship Between Critical Angle and Refractive Index

When light travels from glass to air, and the angle of incidence equals the critical angle (C):

$\sin C = \frac{\sin 90^{\circ}}{<em>{a}\mu</em>{g}} = \frac{1}{<em>{a}\mu</em>{g}}$

Since $\mu = \frac{1}{\sin C}$, then

$\sin C = \frac{1}{\mu}$

Example: If the refractive index of light is 1.5, then

$\sin C = \frac{1}{1.5} = 0.6667$

$C \approx 42^{\circ}$

Total internal reflection occurs if the angle of incidence is greater than 42 degrees.

Reflection vs. Refraction

• Reflection: Waves bounce off the surface; the angle of incidence equals the angle of reflection. Reflection occurs in mirrors.
• Refraction: Waves pass through the surface, changing their speed and direction. Refraction occurs in lenses.

Conditions for Total Internal Reflection

1. Light must be traveling from an optically denser medium to an optically less dense medium.
2. The angle of incidence in the denser medium must be greater than the critical angle.

Uses of Total Internal Reflection

1. Optical fibers: Used in telecommunications and endoscopes. Endoscopes are medical instruments with optical fibers and lenses to see inside the patient’s body.
2. Automatic rain sensors: Used to control windshield wipers.
3. Binoculars: Use glass prisms.
4. Submarine periscopes: Allow submarines to visually search for targets and threats on the surface of the water and in the air while submerged.

Applications of Internal Reflection in Nature

1. Mirage: An optical illusion caused by the refraction of light due to temperature differences between the air near the surface of the road and the air higher up. This makes it appear as if there is a pool of water on the road.
2. Field view of fish underwater: Fish underwater have a full view of everything above the water surface, provided the water is not disturbed.

Calculations/Examples

1. Determine the real depth of a river which appears to be 20m deep when viewed directly from above the surface? (Take refractive index of water = 1.33)

   $\mu = \frac{\text{real depth}}{\text{apparent depth}}$

   $1.33 = \frac{\text{real depth}}{20}$

   Real depth $= 1.33 \times 20 = 26.6m$

2. A ray of light is incident in water at an angle of 20 degrees, from water to a glass plane surface. Calculate the angle of refraction in the glass. (Take $<em>{a}\mu</em>{g} = 1.5$, $<em>{a}\mu</em>{w} = 4/3$)
   $<em>{w}\mu</em>{g} = \frac{<em>{a}\mu</em>{g}}{<em>{a}\mu</em>{w}} = \frac{1.5}{4/3} = \frac{1.5}{1.33} = 1.1278$
   $\frac{\sin i}{\sin r} = 1.1278$
   $\sin r = \frac{\sin 20}{1.1278}$
   $r = \arcsin(\frac{\sin 20}{1.1278}) = 17.67^{\circ}$

3. A ray of light is incident in water at an angle of 30 degrees, from water to air plane surface. Find the angle of refraction in the air. (take refractive index of water =1.33)

   $<em>{w}\mu</em>{a} = \frac{\sin r}{\sin i}$

   $\frac{1}{1.33} = \frac{\sin 30}{\sin r}$

   $\sin r = 1.33 \times \sin 30$

   $\sin r = 1.33 \times 0.5 = 0.665$

   $r = \arcsin(0.665) = 41.68^{\circ}$

4. Determine the critical angle for light traveling from water to air. (take refractive index of water = 4/3)

   $\mu = \frac{1}{\sin C}$

   $\frac{4}{3} = \frac{1}{\sin C}$

   $\sin C = \frac{3}{4} = 0.75$

   $C = \arcsin(0.75) = 48.59^{\circ}$

5. The velocities of light in air and glass are $3.0 \times 10^8 m/s$ and $1.8 \times 10^8 m/s$ respectively. Calculate the sine of the angle of incidence that will produce an angle of refraction of 30 degrees for a ray of light incident on glass.

   $<em>{a}\mu</em>{g} = \frac{v<em>{air}}{v</em>{glass}} = \frac{\sin i}{\sin r}$

   $\frac{3.0 \times 10^8}{1.8 \times 10^8} = \frac{\sin i}{\sin 30}$

   $\sin i = \frac{3.0 \times 10^8}{1.8 \times 10^8} \times \sin 30$

   $\sin i = \frac{3.0}{1.8} \times 0.5$

   $\sin i = 0.833$

6. An electromagnetic wave of frequency $5.0 \times 10^{14} Hz$ is incident on the surface of water of refractive index 4/3. Taking the speed of wave in air as $3.0 \times 10^8 m/s$, calculate the wavelength of the wave in water.

   $f = 5.0 \times 10^{14} Hz$

   $\lambda<em>{air} = \frac{v</em>{air}}{f} = \frac{3.0 \times 10^8}{5.0 \times 10^{14}} = 6.0 \times 10^{-7} m$

   $<em>{a}\mu</em>{w} = \frac{\lambda<em>{air}}{\lambda</em>{water}}$

   $\frac{4}{3} = \frac{6.0 \times 10^{-7}}{\lambda_{water}}$

   $\lambda_{water} = \frac{3}{4} \times 6.0 \times 10^{-7} = 4.5 \times 10^{-7} m$

7. A ray of light is incident at an angle of 30 degrees at an air-glass interface.

   i) Draw a ray diagram to show the deviation of the ray in the glass.

   ii) Determine the angle of deviation (refractive index of glass= 1.5)

   Solution:

   $\mu = \frac{\sin i}{\sin r}$

   $1.5 = \frac{\sin 30}{\sin r}$

   $\sin r = \frac{\sin 30}{1.5} = \frac{0.5}{1.5} = 0.333$

   $r = \arcsin(0.333) = 19.45^{\circ}$

   Angle of deviation (d) = i - r

   $d = 30 - 19.45 = 10.55^{\circ}$

8. A rectangular prism of thickness 12cm is placed on a mark on a piece of paper resting on a horizontal bench, if the refractive index of the material of the prism is 1.5, calculate the apparent displacement of the mark

   Real depth = the thickness = 12cm, $\mu = 1.5$

   $\mu = \frac{\text{real depth}}{\text{apparent depth}}$

   $1.5 = \frac{12}{\text{apparent depth}}$

   Apparent depth = $\frac{12}{1.5} = 8 cm$

   Apparent displacement = real depth – apparent depth

   = 12 – 8 = 4cm