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Light is incident from a medium into air at two possible angles of incidence (A) 20° and (B) 40°. In the medium light travels 3.0 cm in 0.2 ns. The ray will :
First, we need to determine the refractive index of the medium. The problem states that light travels 3.0 cm in 0.2 ns in the medium. Convert the distance to meters: 3.0 cm = 0.03 m. Convert the time to seconds: 0.2 ns = 0.2 × 10⁻⁹ s = 2 × 10⁻¹⁰ s.
The speed of light in the medium, $$ v $$, is given by distance divided by time:
$$ v = \frac{0.03}{2 \times 10^{-10}} = \frac{3 \times 10^{-2}}{2 \times 10^{-10}} = \frac{3}{2} \times 10^{-2 - (-10)} = \frac{3}{2} \times 10^{8} = 1.5 \times 10^{8} \text{m/s} $$
The speed of light in vacuum (or air) is $$ c = 3 \times 10^{8} \text{m/s} $$. The refractive index $$ n $$ of the medium is:
$$ n = \frac{c}{v} = \frac{3 \times 10^{8}}{1.5 \times 10^{8}} = \frac{3}{1.5} = 2 $$
Since light is going from the medium to air, total internal reflection occurs if the angle of incidence is greater than the critical angle $$ \theta_c $$. The critical angle is given by:
$$ \sin \theta_c = \frac{n_{\text{air}}}{n_{\text{medium}}} = \frac{1}{2} = 0.5 $$
Thus, $$ \theta_c = \sin^{-1}(0.5) = 30^\circ $$.
Now, analyze the two cases:
Evaluating the options:
Hence, the correct answer is Option B.
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