Total internal reflection in triangular prism with n=2.56

[GingerBread27]

A corner reflector is to be made from a triangular prism with index of refraction n=2.56, as shown in the diagram on your assignment. What is the maximum angle (in °), with respect to the normal to the front surface of the prism, (theta), such that total reflection will occur?

Ok this must be easy since it is one of the first problems in my assignment , but I'm not getting it, any ideas?

 

[Andrew Mason]

A corner reflector is to be made from a triangular prism with index of refraction n=2.56, as shown in the diagram on your assignment. What is the maximum angle (in °), with respect to the normal to the front surface of the prism, (theta), such that total reflection will occur?

Ok this must be easy since it is one of the first problems in my assignment , but I'm not getting it, any ideas?

I think they want you to use Snell's law:

$$n_{air}sin\theta_1 = n_{glass}sin\theta_2$$ and find $\theta_1$ where $\theta_2 > 45\deg$

If $\theta_2 > 45\deg$, the light will not strike both sides of the prism and will not be reflected back in the direction of the incident ray.

AM

 

[thepatientmental]

The maximum angle (theta) can be calculated using the formula for critical angle, which is given by sin(theta) = 1/n, where n is the index of refraction. In this case, n=2.56, so the maximum angle is sin(theta) = 1/2.56 = 0.3906. To find the angle in degrees, we can use inverse sine function, so theta = sin^-1(0.3906) = 22.5°. This means that any angle greater than 22.5° with respect to the normal to the front surface of the prism will result in total internal reflection. This is because when the angle of incidence is equal to or greater than the critical angle, the light will be completely reflected back into the prism instead of being refracted out. This property of total internal reflection is used in many optical devices, such as fiber optics and corner reflectors, to efficiently redirect light.