In the figure shown, for an angle of incidence 45, at the top surface, what is the minimum refractive index needed for total internal reflection at vertical face AD?

Question

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Answer 1

To find:
Minimum R.I
so that TIR
takes place at AD=?
For TIR
$90 - t = θ_{c}$
Applying snesll's Law at Ab
$u_{1} sin i = u_{2} sin r$
$1 \times sin 45 = u sin r$
$u sin t = \frac{1}{2}$ ............(1)
Applying snell's law at AD
$u sin (90 - t) = 1 \times sin 90$
$u = \frac{1}{cos r}$ ............(2)
From eq (1) and (2)
$tan r = \frac{1}{2}$
from eq (1)
$u . \frac{1}{3} = \frac{1}{2} \Rightarrow u = \frac{3}{2}$

Answer 2

To find:

Minimum R.I

so that TIR

takes place at AD=?

For TIR

90 - t = θ_{c}

Applying snesll's Law at Ab

u_{1} sin i = u_{2} sin r

1 \times sin 45 = u sin r

u sin t = \frac{1}{2}

............(1)

Applying snell's law at AD

u sin (90 - t) = 1 \times sin 90

u = \frac{1}{cos r}

............(2)

From eq (1) and (2)

tan r = \frac{1}{2}

from eq (1)

u . \frac{1}{3} = \frac{1}{2} \Rightarrow u = \frac{3}{2}