A triangular glass prism with an apex angle of Φ has an index of refraction n (see figure below). What is the smallest angle of incidence θ1 for which a light ray can emerge from the other side?

+3 votes

A triangular glass prism with an apex angle of Φ has an index of refraction n (see figure below). What is the smallest angle of incidence θ1 for which a light ray can emerge from the other side? (Use the following as necessary: Φ and n.)
θ1 =

image
asked Apr 7, 2012 in Physics by awesome ~Expert~ (1,479 points)
    

1 Answer

+2 votes
 
Best answer

Using snell's law, we know that at the first refraction we have:

1*sinθ1 = n*sinθ2

The critical angle at the second surface is given by:

n * sinθ3 = 1, so

θ3 = inverse sin( 1 / n )

Using the geometry of the triangle, we can conclude that:

(90° − θ2) + Φ + (90° − θ3) = 180°,

so

θ2 = Φ − θ3.

So to have

θ3 = inverse sin( 1 / n ) and to avoid total internal relection at the second surface, it is necessary that:

θ2 = Φ − inverse sin( 1 / n )

plugging back in the first relation we get that:

sinθ1 = n*sin(Φ − inverse sin( 1 / n ))

and so

θ1 = inverse sin (n*sin(Φ − inverse sin( 1 / n )) )

simplifying using trig identites we get that:

answered Apr 7, 2012 by pokemonmaster ~Expert~ (3,856 points)
selected Apr 11, 2012 by awesome
Thanks! :)

Related questions




...