Question

3. (a) Assume that a light ray incident from air on a spherical raindrop of radius R at angle 0 measured with respect to the surface normal undergoes one internal reflection before leaving the raindrop, as shown below. refraction Internal Incident reflection 8 ray R refraction Use the area above to show that if e0 and e 0t, then the total deflection angle & 8+88 of the light ray is (in radians): 8 2(0,-0)+- 20, The deflection angle & is the angle between the incident ray and the outgoing ray, and e, is the angle of refraction for the light ray incident from air and entering water. (b) Eliminate e from the equation for 8 and express & in terms of 0, nair, and nwater (c) If 60°, is the internal reflection a total internal reflection?

Answer 1

refraction 0 Incident ray Internal reflection refraction

name the point first incident, internal reflection, last exit as a,b,c

first deflection :  $\delta$₁ = $\theta$ _1i - $\theta$ _1t = $\theta$ _i - $\theta$ _t  

second deviation :  $\delta$₂ = 180 - ($\theta$_2i +  $\theta$_2r )

using reflection law $\theta$ _2i =  $\theta$_2r

triangle oab is isoceless $\theta$ _1t =  $\theta$_2i  

$\delta$₂ = 180 - 2$\theta$_t

third deviation : $\delta$ ₃ = $\theta$ _3t -  $\theta$_3i

using the triangle rule as earlier   $\theta$_3i = $\theta$ _1t =$\theta$_t  

$\theta$_3t = $\theta$ _1i = $\theta$ _i

$\delta$₃ = $\theta$ _i -  $\theta$_t  

$\delta$ = $\delta$ ₁ + $\delta$ ₂ + $\delta$ ₃ = 2($\theta$_i -  $\theta$_t  ) + ($\pi$ - 2$\theta$_t )  

b) Sin$\theta$_i /Sin$\theta$_t = n_water /n_air

$\theta$_t = Sin^-1(n_air Sin$\theta$_i /n_water )

$\delta$ =  2$\theta$_i + $\pi$ - 4$\theta$_t = 2$\theta$_i + $\pi$ - 4Sin^-1(n_air Sin$\theta$_i /n_water )

c) $\theta$ _i = 60

Sin$\theta$_t= Sin(60)/ n_w

$\theta$_t = 40.6 deg

inside the drop angle of incidence = 40.6 deg. It is not TIR. For TIR the angle of incidence must be more than 49 deg.