Question

as light goes from a material with a low index of refraction (like air) to a material with a high index of refraction (like water), the light is bent

towards the normal

away from the normal

totally reflected

totally refracted

question 4 (1 point)
what is the maximum angle of incidence going from water to air that still has an angle of refraction?

70

60

25

49

Explanation:

First Question (Light Refraction Direction)

When light travels from a medium with a lower refractive index (air) to a higher one (water), according to the law of refraction (Snell's Law), the light bends towards the normal. Total reflection occurs when going from high to low index at a critical angle, and "totally refracted" is not a standard term. Bending away happens when going from high to low index.

Step1: Recall Critical Angle Formula

The critical angle \( \theta_c \) is given by \( \sin\theta_c=\frac{n_2}{n_1} \), where \( n_1 \) is the refractive index of the medium the light is coming from (water, \( n_1 = 1.33 \)) and \( n_2 \) is the refractive index of the medium it's going to (air, \( n_2 = 1 \)).

Step2: Calculate Critical Angle

Substitute \( n_1 = 1.33 \) and \( n_2 = 1 \) into the formula: \( \sin\theta_c=\frac{1}{1.33}\approx0.7519 \). Then, \( \theta_c=\arcsin(0.7519)\approx48.8^\circ\approx49^\circ \). This is the maximum angle of incidence in water (optically denser medium) to air (optically rarer medium) where refraction still occurs (beyond this, total internal reflection happens).

Answer:

A. towards the normal

Question 4 (Critical Angle for Water to Air)