The speed of light is 1.75 times larger in material A than it is inmaterial B. Determine the ratio nA/nB of the refractive indices ofthese materials.

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In Example 33.11, the water-glass interface is horizontal. If instead this interface were tilted 15.0° above the horizontal, with the right side higher than the left side, what would be the angle from the vertical of the ray in the glass? (The ray in the water still makes an angle of 60.0° with the vertical).

EXAMPLE 33.11: Reflection and refraction

In Fig 33.11, material a is water and material b is a glass with an index of refraction 1.52. If the incident ray makes an angle of 60.0° with the normal, find the directions of the reflected and refracted rays.

SOLUTION

IDENTIFY: This is a problem in geometric optics. We are given the incident angle and the index of refraction of each material, and we need to find the reflected and refracted angles.

SET-UP: Figure 33.11 shows the rays and angles for this situation. The target variables are the reflected angle θ_r and the refracted angle θ_b. Since n_b is greater than n_a, the refracted angle must be smaller than the incident angle θ_a; this is shown in the figure.

EXECUTE: According to Eq 33.2, the angle the reflected ray makes with the normal is the same as that of the incident ray so θ_r= θ_a=60.0°.

To find the direction of the refracted ray, we use Snell’s law, Eq 33.4, with n_a=1.33, n_b=1.42, and θ_a=60.0°. We find:

EVALUATE: The second material has a larger refractive index than the first, just like the situation shown in Fig 33.8a. Hence, the refracted ray is bent toward the normal as the wave slows down upon entering the second material and θ_b< θ_a.

Figure 33.11. Reflection and refraction of light passing from water to glass

Figure 33.8. Reflection and refraction in three cases. (a) Material b has a larger index of refraction than material a. (b) (a) Material b has a smaller index of refraction than material a. (c) The incident light ray is normal to the interface between the materials.