Question

Light enters at an angle of incidence in a transparent rod of refractive index n. For what value of the refractive index of the material of the rod the light once entered into it will not leave it through its lateral face whatsoever be the value of angle of incidence?(a) n > √2(b) n = 1(c) n = 1.1(d) n = 1.3

Answer 1

Answer:

Explanation:

Let a ray of light enter at A and the refracted beam is AB. Thus, the incident will be at an angle q.

For no refraction at the lateral face, q > C or, sin q > sin C

But q + r = 90° Þ q = (90° - r)

Therefore sin (90° - r) > sin C or cos r > sin C

The substitution for cos r can be found from Snell's law. Thus, according to the from Snell's law -

n = sini/sinr = sinr = sini/n

= cosr = √1−sin²r

= √1−sin²i/n²

= √1−sin²i/n² > sin C

= 1−sin²i/n² > sin² C

Therefore, the maximum value of sin i is 1.

So, the value of angle of incidence is n² >2 or n>√2