If the critical angle for total internal reflection from a medium to vacuum is $45^{\circ}$, the velocity of l

Question

If the critical angle for total internal reflection from a medium to vacuum is $45^{\circ}$, the velocity of light in the medium is:

Accepted Answer

1. **Critical Angle Relation:** The refractive index ($\mu$) of a medium is related to the critical angle ($C$) for total internal reflection by the formula: $\mu = \frac{1}{\sin C}$.
2. **Calculate Refractive Index:** Given $C = 45^{\circ}$.
   $$ \mu = \frac{1}{\sin 45^{\circ}} = \frac{1}{1/\sqrt{2}} = \sqrt{2} $$
3. **Velocity Relation:** The velocity of light in a medium ($v$) is related to the speed of light in vacuum ($c$) and the refractive index ($\mu$) by: $v = \frac{c}{\mu}$.
4. **Calculation:** Substituting $c = 3 	imes 10^8 	ext{ m/s}$ and $\mu = \sqrt{2}$:
   $$ v = \frac{3 	imes 10^8}{\sqrt{2}} 	ext{ m/s} $$
5. **Conclusion:** The velocity of light in the medium is $\frac{3}{\sqrt{2}} 	imes 10^8 	ext{ m/s}$.

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