Light with a wavelength of 589 nm in a vacuum strikes the surface of an unknown liquid at an angle of 31.2° with respect to the normal to the surface.If the light travels at a speed of 1.97 × 10⁸ m/s through the liquid,what is the angle of refraction?
A)19.9°
B)26.1°
C)34.2°
D)39.3°
E)51.9°

Question

Quizplus · Accepted Answer

To find the angle of refraction, we use Snell's Law, which is $n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$, where $n_1$ and $n_2$ are the refractive indices of the first and second mediums respectively, and $\theta_1$ and $\theta_2$ are the angles of incidence and refraction respectively. The refractive index of a medium is given by $n = \frac{c}{v}$, where $c$ is the speed of light in vacuum (approximately $3.00 \times 10^8$ m/s) and $v$ is the speed of light in the medium.Given that the light travels at a speed of $1.97 \times 10^8$ m/s through the liquid, the refractive index of the liquid is $n_2 = \frac{3.00 \times 10^8}{1.97 \times 10^8} \approx 1.52$. The refractive index of vacuum ($n_1$) is 1, and the angle of incidence ($\theta_1$) is 31.2°.Substituting these values into Snell's Law gives us $1 \times \sin(31.2°) = 1.52 \times \sin(\theta_2)$. Solving for $\theta_2$ gives us an angle of refraction of approximately 19.9°.

Light with a Wavelength of 589 Nm in a Vacuum