critical angle & angle of theta?

Question

19) what is the critical angle for diamond (n= 2.42)?

20) A diver below the surface of a lake notices that no light appears to be coming toward her greater then the angle theta, measured from the vertical. If the index of refraction of water is 1.33, what is the angle of theta??


please help me and explain this to me, I'm having a hard time, thanks

Answer 1

19) arcsin (1/2.42) = 24.4°

20) θ = arcsin (1/1.33) = 48.75°

Answer 2

Remember Snell's law for refraction: n1·sinθ1 = n2·sinθ2, where θ is measured from the normal (line perpendicular to the surface). This tells you that light going from a denser medium to a less dense one is refracted away from the the normal - if n1 > n2, then θ2 > θ1. Now θ1 (the angle in the denser medium) can take any value from 0 to 90°, as can θ2. But there comes a point as θ1 increases, that sinθ2 = 1 and θ2 = 90°, that is, the light is refracted parallel to the surface. Since θ2 can't be more than 90°, the θ1 in this case this marks the limiting angle for light passing through the boundary. that value of θ1 is called the critical angle. You calculate the critical angle by setting sinθ2 to 1, then sinθ1 = n2/n1.

Answer 3

When optical fibre & air are taken ,the former is denser medium and the later is rarer medium.For critical angle the ray approaches from denser medium(incident med.) to rarer medium(refractive med.) i.e, from optical fibre(ni) to air(nr).As refractive index of air is 1,nr is taken as1.