A glass vessel in the shape of a triangular is filled with water, and light is incident normally on the face XY. If the refractive indices for water and glass are $4 / 3$ and $3 / 2$ respectively, total internal reflection will occur at the glass- air surface XY only. What is the minimum value of sin $θ$

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Answer 1

$μ = \frac{1}{sin C}$ [ where u is the refractive index $(\frac{u _{2}}{u _{1}})$ C is critical angle ]
$\Rightarrow \frac{u _{2}}{u _{1}} = \frac{1}{sin C}$

$\Rightarrow \frac{\frac{3}{2}}{\frac{4}{3}} = \frac{1}{sin C}$

$\Rightarrow \frac{3}{2} \times \frac{3}{4} = \frac{1}{sin C}$

$\Rightarrow \frac{9}{8} = \frac{1}{sin C}$

$∴ sin C = \frac{8}{9}$
$C = 62.73 °$ ( approx. )
Incident ray is perpendicular.
Therefore it would not show any refraction or reflection. When incident ray reaches AC.
It would make a $△$ with angle of prism $6 0^{0}$ and with $a c = 180 ° 60 ° + 90 ° = 30 °$
Angle of incidence $= 90 ° - 30 ° = 60 °$
Since incident angle is less than critical angle.

Answer 2

μ = \frac{1}{sin C}

[ where u is the refractive index

(\frac{u _{2}}{u _{1}})

C is critical angle ]

\Rightarrow \frac{u _{2}}{u _{1}} = \frac{1}{sin C}

\Rightarrow \frac{\frac{3}{2}}{\frac{4}{3}} = \frac{1}{sin C}

\Rightarrow \frac{3}{2} \times \frac{3}{4} = \frac{1}{sin C}

\Rightarrow \frac{9}{8} = \frac{1}{sin C}

∴ sin C = \frac{8}{9}

C = 62.73 °

( approx. )

Incident ray is perpendicular.

Therefore it would not show any refraction or reflection. When incident ray reaches AC.

It would make a

△

with angle of prism

6 0^{0}

and with

a c = 180 ° 60 ° + 90 ° = 30 °

Angle of incidence

= 90 ° - 30 ° = 60 °

Since incident angle is less than critical angle.