What Is the Critical Angle Between Glass and Water?

[MYS9]

A light ray traveling through a glass medium, index of refraction1.52, is incident on the boundary between the glass and water, index of refraction 1.33 What is the critical angle for the glass at this boundary? What would the critical angle be if the incident medium was air instead of water?

would i use snell's law? how?

thanks

 

[Chi Meson]

Yes, Snell's law. The critical angle only exists for a wave that is going from the slow medium to the fast medium. The critical angle is the angle of incidence that causes the angle of refraction to be exactly 90 degrees. Set up the formula using "theta 2" as 90 and solve for "theta 1."

 

[M98Ranger]

Yes, you would use Snell's law to solve this problem. The critical angle can be found by using the equation sinθc = n2/n1, where θc is the critical angle, n1 is the index of refraction of the incident medium, and n2 is the index of refraction of the medium that the light is entering.

For this problem, n1 is 1.52 (glass) and n2 is 1.33 (water). Plugging these values into the equation, we get sinθc = 1.33/1.52 = 0.875. To find the critical angle, we need to take the inverse sine of 0.875, which gives us θc = 60.69°.

If the incident medium was air instead of water, n2 would be 1 (since air has an index of refraction of approximately 1). Plugging this into the equation, we get sinθc = 1/1.52 = 0.657. Taking the inverse sine, we get θc = 41.81°.

So, the critical angle for the glass at the boundary with water is 60.69°, but if the incident medium was air, the critical angle would be smaller at 41.81°. This means that the light ray would have a greater chance of being reflected back into the glass when entering from air, compared to when entering from water.

 

[M98Ranger]

Yes, you would use Snell's Law to calculate the critical angle in this scenario. The critical angle is the angle of incidence at which the light ray will be refracted at an angle of 90 degrees, meaning it will travel parallel to the surface instead of passing through it.

To calculate the critical angle, we can use the formula: sin(critical angle) = n2/n1, where n2 is the index of refraction of the incident medium (in this case, water) and n1 is the index of refraction of the medium the light is coming from (in this case, glass).

For this scenario, the critical angle for the glass at the boundary with water would be sin(critical angle) = 1.33/1.52, which gives us a critical angle of approximately 49.9 degrees.

If the incident medium was air instead of water, the critical angle would be different because the index of refraction of air is 1.00. Using the same formula, the critical angle for the glass at the boundary with air would be sin(critical angle) = 1.00/1.52, which gives us a critical angle of approximately 41.8 degrees.

It is important to note that the critical angle will be different for different materials and different combinations of materials. It is also worth mentioning that if the angle of incidence is greater than the critical angle, total internal reflection will occur, meaning the light will be completely reflected back into the incident medium instead of being refracted.