Intensity of Light Through Pentaprism Including Reflections

[Ichigo449]

Homework Statement

A pentaprism is a five-sided reflecting prism used to deviate a beam of light by 90◦ without inverting an image. Pentaprisms are used in the viewfinders of SLR cameras.
(a) What prism angle β is required for a normal-incidence beam from the left to exit the bottom surface at normal incidence?
(b) If all interfaces of the pentaprism are uncoated glass with index n = 1.5, what fraction of the intensity would get through this system for a normal incidence beam? Compute for p-polarized light, and include transmission through the first and final surfaces as well as reflection at the two interior surfaces.

Homework Equations

Fresnel Equations

The Attempt at a Solution

a) I really am lost at how to find the prism angle. I know it should come from an angle of deviation calculation but cannot set one up.
b) The overall transmission through the device is the repeated application of the Fresnel equations at each point mentioned: the transmittance from air to glass at normal incidence: 96 % I_o. Then the reflectance at the angle required for total internal reflection twice (this is 41 degrees), and finally transmittance from glass to air again.
Is solution process correct?

 

[blue_leaf77]

Do you have any related picture about the prism. I find it difficult to find a reference about which corner is referred to by "prism angle" in a pentaprism?

 

[Ichigo449]

The image is at https://photos.google.com/album/AF1QipMhPZIXS9qC2Go96kqaPS2d8V_d_o71g1_-zRoP

 

[blue_leaf77]

The image is at https://photos.google.com/album/AF1QipMhPZIXS9qC2Go96kqaPS2d8V_d_o71g1_-zRoP

Couldn't access that link.

 

[Ichigo449]

The question is from http://optics.byu.edu/BYUOpticsBook_2015.pdf Problem 3.5

 

[blue_leaf77]

Based on that picture, what is $2\theta_i$ equal to?

 

[Ichigo449]

Since the light rays form a right triangle and the Law of Reflection holds 2\theta_i =\pi/4.

 

[blue_leaf77]

Since the light rays form a right triangle and the Law of Reflection holds 2\theta_i =\pi/4.

Yes, that's right. With this knowledge alone and some creativity in trigonometry, you can determine $\beta$.

 

[Ichigo449]

Could you give me a hint for what triangles to consider?

 

[blue_leaf77]

Consider one of the reflection points inside the prism. What is the angle subtended by the incoming (or reflected) ray with the prism's side?

 

[Ichigo449]

I can't see a subtended angle for one reflection point, but $\beta$ is the subtended angle from both reflection points.

 

[blue_leaf77]

By "reflection point", I mean the spot where the ray hits the inner side of the prism at an angle $\theta_i$ and reflects off also at the same angle. What I suggested is that you calculate the angle subtended by the ray hitting this reflection point and the prism's side, i.e. the complementary angle of $\theta_i$.

 

[Ichigo449]

Thank you. I think I understand the process now. The Law of Reflection forces the complementary angles at both points to be equal, and these two angles are included in the triangle I mentioned so $\beta = \pi/4$ because each complementary angle is $3\pi/8$.

 

[blue_leaf77]

so $\beta = \pi/4$

Yes that's right.

 

[Ichigo449]

So for part b) am I applying the Fresnel equations correctly?

 

[blue_leaf77]

Yes.

 

[Ichigo449]

So I went through the calculations and 4% is lost on first transmission, 0.0199 on each reflection and 4% is again lost on exiting. This is only a net loss in intensity of 8.0398%. So I made a mistake somewhere. For the reflections at oblique incidence I used that $R = r_s^2 = tan^2 (\theta_t -\theta_i) /tan^2 (\theta_t +\theta_i)$