Finding the incidence angle for refraction

[SirBerr]

Homework Statement

I am to find the angle of incidence of a light ray before it undergoes refraction. Light is refracted in the second medium at an angle of 76 degrees measured from the normal. The second index of refraction is 1.386. The index of refraction for the initial or first medium is 1.336.

Homework Equations

Snell's Law n1sin(theta 1) = n2 sin (theta 2)

The Attempt at a Solution

I worked through Snell's Law and found that sin ( theta one) is 1.006 . When I attempt to take the inverse SIN for the angle, I get domain error because the value is greater than 1.

This can't be internal reflection because the second medium has a higher index of refraction than the first?

All help is appreciated! Thanks

 

[TSny]

Although this is not total internal reflection, the situation here is related to the critical angle.

If a ray of light refracts going from material 1 to material 2, then that ray should be "reversible". That is, light should be able to travel the same path in the reversed direction from material 2 back into material 1.

So, for the reversed path in your problem, the ray would start in material 2 and have an angle of incidence of 76 degrees. How is 76 degrees related to the critical angle for going from material 2 to material 1?

Another way to look at the problem is to consider the light going from material 1 into material 2 with greater and greater angles of incidence in material 1. As you increase the angle of incidence in material 1, what happens to the angle of refraction in material 2? What would you let the angle of incidence be to find the greatest possible angle of refraction in material 2? How does this maximum angle of refraction compare with 76 degrees?

 

[SirBerr]

Although this is not total internal reflection, the situation here is related to the critical angle.

If a ray of light refracts going from material 1 to material 2, then that ray should be "reversible". That is, light should be able to travel the same path in the reversed direction from material 2 back into material 1.

So, for the reversed path in your problem, the ray would start in material 2 and have an angle of incidence of 76 degrees. How is 76 degrees related to the critical angle for going from material 2 to material 1?

Another way to look at the problem is to consider the light going from material 1 into material 2 with greater and greater angles of incidence in material 1. As you increase the angle of incidence in material 1, what happens to the angle of refraction in material 2? What would you let the angle of incidence be to find the greatest possible angle of refraction in material 2? How does this maximum angle of refraction compare with 76 degrees?

I noticed you left reversible in quotations. Why must we address a reversible ray when the ray is not physically going from a high medium to a lower medium of index?

I can see what you're saying as the incidence angle is increased, I even did some online simulators, but I cannot grasp as to why this occurs. I thought this type of reflection only occurred again when going from a high index to a low index.

 

[TSny]

I noticed you left reversible in quotations. Why must we address a reversible ray when the ray is not physically going from a high medium to a lower medium of index?

If a ray of light is possible in one direction, then the reversed ray must also be possible. (Both rays will satisfy Snell's law). See the attached picture. So, if you can show that the reversed ray is impossible, that means the original ray is impossible. That's why you can't get an answer with your calculator.

I can see what you're saying as the incidence angle is increased, I even did some online simulators, but I cannot grasp as to why this occurs. I thought this type of reflection only occurred again when going from a high index to a low index.

We're not talking about reflection here, just refraction in going from medium 1 to medium 2. There is a greatest value that θ₂ can be. If 76^o is greater than that maximum possible angle of refraction, then it's impossible for a ray to be refracted at 76^o. So, that's another way to see why your calculator is giving an error.

 

[Nickie]

I would first commend you for attempting to solve the problem using Snell's Law. However, it seems like there may be a calculation error in your attempt to find the angle of incidence. The sine of an angle cannot be greater than 1, so it is possible that you made a mistake in your calculations.

To find the angle of incidence, we can rearrange Snell's Law to solve for theta 1:

theta 1 = sin^-1 (n2/n1 * sin(theta 2))

Plugging in the values given in the problem, we get:

theta 1 = sin^-1 (1.386/1.336 * sin(76)) = 55.8 degrees

This angle of incidence makes sense based on the information given in the problem. It is less than 76 degrees, which means the light ray is being bent towards the normal as it enters the second medium with a higher index of refraction. If you are still getting a domain error when taking the inverse sine, I would recommend checking your calculations again or using a calculator that can handle values greater than 1.

In terms of the possibility of internal reflection, it is important to note that the critical angle for total internal reflection is when the angle of incidence is equal to the inverse sine of the ratio of the two indices of refraction. In this case, that would be when the angle of incidence is equal to 54.7 degrees. Since our calculated angle of incidence is slightly higher than this, it is possible that there could be some internal reflection happening, but it would not be the dominant effect. The majority of the light would still be refracted into the second medium.

In conclusion, it is important to double check your calculations and make sure you are using the correct equations and values. In this case, the angle of incidence can be found using Snell's Law and is approximately 55.8 degrees.