How Do You Calculate Lateral Displacement in Refraction Problems?

In summary, the problem involves calculating the lateral displacement between an incident and emergent ray of light passing through a glass plate with a thickness of 2.40 cm and an index of refraction of 1.80. The formula of n_1sin(theta_1) = n_2sin(theta_2) is used to calculate the angle of refraction, and trigonometry is needed to determine the distance between the two rays. A proper ray diagram can be drawn to compare the angles and derive an equation for the displacement.
  • #1
sghaussi
33
0
Hello! I'm having difficulty solving this problem, I was wondering if you can help me solve it.

Light is incident in air at an angle on the upper surface of a transparent plate, the surfaces of the plate being plane and parallel to each other.

A ray of light is incident at an angle of 66.0 on one surface of a glass plate 2.40 cm thick with an index of refraction 1.80. The medium on either side of the plate is air. Find the lateral displacement between the incident and emergent rays.


I know I am supposed to use the formula: n_1sin(theta_1) = n_2sin(theta_2)
index of refraction of air = 1.00


However I don't know how to approach this problem. I'm not asking for the answer, just maybe a few steps on how to approach this type of problem? And will I need to use trig to find the distance between the incident and emergent rays?


Thank you,
Sahar
 
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  • #2
Draw yourself a picture of what's going on. When the light hits the glass, it's refracted at an angle given by Snell's law. Calculate the angle, then use a bit of trig to figure out where the light exits the glass (compared to the where it enters). Then figure out where the light would exit the glass if it weren't refracted. Compare.
 
  • #3
And remember that Snell's Law calculates the angle with respect to the normal line. Do they want the angle between them or the distance between them?
 
  • #4
You need to calculate the lateral displacement between the incident and emergent rays. Note that the lateral displacement is the perpendicular distance between the two rays. (Not just the distance between the exit points.)
 
  • #5
Draw a proper ray diagram, you'll find two triangles with the same base, in terms of the first and second incident angles, which can be compared to give a value for s, displacement. Simply derive the equation for s, it'll be in terms of the thickness and angle 1 and 2 (first incident and emerging rays).
 

Related to How Do You Calculate Lateral Displacement in Refraction Problems?

1. What is the difference between reflection and refraction?

Reflection is the bouncing back of light or other waves when they encounter a boundary between two different mediums. Refraction is the bending of light or other waves when they pass through a boundary between two different mediums.

2. How do reflection and refraction affect the speed of light?

Reflection has no effect on the speed of light, as the light bounces back at the same speed it entered. Refraction, on the other hand, can change the speed of light by slowing it down or speeding it up, depending on the properties of the two mediums it is passing through.

3. What causes light to reflect and refract?

Light reflects and refracts due to the change in the speed of the wave as it passes through different mediums. This change in speed causes the wave to change direction, resulting in reflection or refraction.

4. How does the angle of incidence affect reflection and refraction?

The angle of incidence is the angle at which the light wave strikes the boundary between two mediums. When this angle changes, it can affect the angle at which the light is reflected or refracted. For example, if the angle of incidence is increased, the angle of reflection will also increase, while the angle of refraction will decrease.

5. What are some real-world applications of reflection and refraction?

Reflection and refraction have many practical applications in everyday life. Some examples include the use of mirrors for reflection in telescopes, microscopes, and other optical devices, and the use of lenses for refraction in cameras, glasses, and binoculars. They are also important in the study of light and optics, and in industries such as fiber optics and telecommunications.

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