Consider a light ray traveling between air and a diamond cut in the shape shown in the figure below.

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Consider a light ray traveling between air and a diamond cut in the shape shown in the figure below. (Take θ = 37.2°.)

image
(a) Find the critical angle for total internal reflection for light in the diamond incident on the interface between the diamond and the outside air. (The index of refraction for diamond is 2.419, and the index of refraction for air is 1.000.)
°

(b) Consider the light ray incident normally on the top surface of the diamond as shown in the figure above. Show that the light traveling toward point P in the diamond is totally reflected. (Do this on paper. Your instructor may ask you to turn in this work.)

(c) Suppose the diamond is immersed in water. What is the critical angle at the diamond–water interface? (The index of refraction for water is 1.333.)
°

(d) When the diamond is immersed in water, does the light ray entering the top surface in the figure undergo total internal reflection at P?
    

Explain.

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(e) If the light ray entering the diamond remains vertical as shown in the figure above, which way should the diamond in the water be rotated about an axis perpendicular to the page through O so that light will exit the diamond at P?
    

(f) At what angle of rotation in part (e) will light first exit the diamond at point P?
° 

 

asked Apr 8, 2012 in Physics by Joey33 ~Expert~ (1,216 points)
    

1 Answer

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Best answer

Part A:

For more information regarding internal reflection check out this page on wikipedia:

http://en.wikipedia.org/wiki/Total_internal_reflection

But basicly here's some background info before we answer the question:

The critical angle is the angle of incidence above which total internal reflection occurs. The angle of incidence is measured with respect to the normal at the refractive boundary (see diagram illustrating Snell's law). Consider a light ray passing from glass into air. The light emanating from the interface is bent towards the glass. When the incident angle is increased sufficiently, the transmitted angle (in air) reaches 90 degrees. It is at this point no light is transmitted into air. The critical angle \theta_c is given by Snell's law,

n1sin θ1 = n2 sin θc       (*since index of refraction of air is 1, we are left with only sin θ1 on the left side)

Furthermore, when finding the critical angle θ1 must equal 90, we are left with 1 on the left hand side:

n2 sin θc = n1

so

θc = inverse sin( n1 / n2 )

plug in n1 and n2 to solve.

 

Part C is essentially the same as Part A, only this time n1 = 1.333 rather than 1.

Once again use this formula to solve

θc = inverse sin( n1 / n2 )

Part D:

The angle of incidence is 37.2°. So, Yes. In this case, the angle of incidence is just larger than the critical angle, so the light ray again undergoes total internal reflection at P.

 

Part E, it would need to be rotated clockwise

 

Part F,

I am not sure of the solution, if anyone else knows please comment! thanks.

 

 

answered Apr 8, 2012 by pokemonmaster ~Expert~ (3,856 points)
selected Apr 8, 2012 by Joey33

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