Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

$1.5 \times 10^8 \text{ m/s}$

$\frac{3}{\sqrt{2}} \times 10^8 \text{ m/s}$

$\sqrt{2} \times 10^8 \text{ m/s}$

$3 \times 10^8 \text{ m/s}$

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

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