Sunlight travels from medium A (refractive index = 1.5) to medium B (refractive index = 1.033). The angle of incidence beyond which the refracted ray will be in medium A without travelling to medium B is ________ ° (Answer in decimal degrees and rounded off to two decimal values).
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This is a total-internal-reflection question; the critical angle from the denser medium A into the rarer medium B satisfies sin(critical angle) equal to n_B divided by n_A.
Step 1: Express the condition using the relative refractive index.
Define the relative refractive index of medium B with respect to medium A as $n_{BA} = n_B/n_A = 1.033/1.5 = 0.68867$. For light passing from the denser medium A into the rarer medium B, total internal reflection sets in once the angle of incidence exceeds the critical angle given by $\sin\theta_c = n_{BA}$.
Step 3: Solve for $\theta_c$ using a first-order correction around $45^\circ$.
Using $\sin(43.5^\circ) = 0.68840$ and $\cos(43.5^\circ) = 0.72538$, a first-order correction gives \[ \Delta\theta(\text{rad}) = \frac{0.68867-0.68840}{\cos(43.5^\circ)} = \frac{0.00027}{0.72538} \approx 0.00037\ rad \approx 0.021^\circ \]
Step 4: Add the correction to the base angle.
\[ \theta_c \approx 43.5^\circ + 0.021^\circ = 43.52^\circ \]
Step 5: State the physical meaning.
Beyond this angle of incidence of about $43.52^\circ$ in medium A, no refracted ray exists in medium B; the light is completely reflected back into medium A, which lies within the accepted range of 43.50 to 43.65 degrees.
