A beam of light in glass \(\left(n_1=1.5\right)\) strikes an interface with water \(\left(n_2=4 / 3\right)\). The critical angle at which total internal reflection takes place is most nearly (A) \(0^{\circ}\) (B) \(48.6^{\circ}\) (C) \(62.7^{\circ}\) (D) \(90^{\circ}\) (E) Total internal reflection cannot take place

Snell's Law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the refractive indices of the two media: \(n_1 \sin{\theta_1} = n_2 \sin{\theta_2}\) Here, \(n_1\) is the refractive index of glass (\(1.5\)), \(n_2\) is the refractive index of water (\(4/3\)), \(\theta_1\) is the angle of incidence, and \(\theta_2\) is the angle of refraction.

The critical angle, \(θ_c\), is the angle of incidence at which the angle of refraction is equal to \(90°\). In this case, \(\theta_1 = \theta_c\) and \(\theta_2 = 90°\). Plugging these values into Snell's Law: \(n_1 \sin{\theta_c} = n_2 \sin{90°}\) Since \(\sin{90°} = 1\), the equation simplifies to: \(n_1 \sin{\theta_c} = n_2\) Now we can isolate the critical angle: \(\sin{\theta_c} = \frac{n_2}{n_1}\) \(\theta_c = \arcsin\left(\frac{n_2}{n_1}\right)\) Our values are \(n_1=1.5\) and \(n_2=4/3\). Substitute these values into the equation: \(\theta_c = \arcsin\left(\frac{4/3}{1.5}\right) = \arcsin\left(\frac{4}{6}\right)\) \(\theta_c = \arcsin(\frac{2}{3})\) Now, we find the value of the angle in degrees: \(\theta_c \approx 41.8^{\circ}\)

From the given options, the angle most nearly equal to the critical angle we calculated is: (B) \(48.6^{\circ}\) Thus, the closest answer to the critical angle at which total internal reflection takes place is (B) \(48.6^{\circ}\).