Trigonometry Snell’s Law Help (page 2)
From Low To High
When a ray of light encounters a boundary between two substances having different indices (or indexes) of refraction, the extent to which the ray is bent can be determined according to an equation called Snell’s law .
Look at Fig. 10-4. Suppose B is a flat boundary between two media M r and M s , whose indices of refraction are r and s, respectively. Imagine a ray of light crossing the boundary at point P, as shown. The ray is bent at the boundary whenever the ray does not lie along a normal line, assuming the indices of refraction, r and s, are different.
Suppose r < s; that is, the light passes from a medium having a relatively lower refractive index to a medium having a relatively higher refractive index. Let N be a line passing through point P on B, such that N is normal to B at P. Suppose R is a ray of light traveling through M r that strikes B at P. Let θ be the angle that R subtends relative to N at P. Let S be the ray of light that emerges from P into M s . Let ø be the angle that S subtends relative to N at P. Then line N, ray R, and ray S all lie in the same plane, and ø ≤ θ . (The two angles θ and ø are equal if and only if ray R strikes the boundary at an angle of incidence of 0°, that is, along line N normal to the boundary at point P. ) The following equation holds for angles θ and ø in this situation:
sin ø /sin = ø = r/s
The equation can also be expressed like this:
s sin ø = sin r sin ø
From High To Low
Refer to Fig. 10-5. Again, let B be a flat boundary between two media M r and M s , whose absolute indices of refraction are r and s, respectively. In this case imagine that r > s; that is, the ray passes from a medium having a relatively higher refractive index to a medium having a relatively lower refractive index. Let N, B, P, R, S, θ , and ø be defined as in the previous example. Then line N, ray R, and ray S all lie in the same plane, and θ ≤ ø . (The angles θ and ø are equal if and only if R is normal to B. ) Snell’s law holds in this case, just as in the situation described previously:
sin ø /sin ø = r/s
s sin ø = r sin θ
Determining The Critical Angle
In the situation shown by Fig. 10-5, the light ray passes from a medium having a relatively higher index of refraction, r, into a medium having a relatively lower index, s. Therefore, s < r. As angle θ increases, angle ø approaches 90°, and ray S gets closer to the boundary plane B. When θ, the angle of incidence, gets large enough (somewhere between 0 and 90°), angle ø reaches 90°, and ray S lies exactly in plane B. If angle θ increases even more, ray R undergoes total internal reflection at the boundary plane B. Then the boundary acts like a mirror.
The critical angle is the largest angle of incidence that ray R can subtend, relative to the normal N, without being reflected internally. Let’s call this angle θ c . The measure of the critical angle is the arcsine of the ratio of the indices of refraction:
θ c = arcsin ( s/r )
Snell's Law Practice Problems
Suppose a laser is placed beneath the surface of a freshwater pond. The refractive index of fresh water is approximately 1.33, while that of air is close to 1.00. Imagine that the surface is perfectly smooth. If the laser is aimed upwards so it strikes the surface at an angle of 30.0° relative to the normal, at what angle, also relative to the normal, will the beam emerge from the surface into the air?
Envision the situation in Fig. 10-5 “upside down.”
Then M r is the water and M s is the air. The indices of refraction are r = 1.33 and s = 1.00. The measure of angle θ is 30.0°. The unknown is the measure of angle ø Use the equation for Snell’s law, plug in the numbers, and solve for ø . You’ll need a calculator. Here’s how it goes:
What is the critical angle for light rays shining upwards from beneath a freshwater pond?
Use the formula for critical angle, and envision the scenario of Problem 10-2, where the laser angle of incidence, θ, can be varied. Plug in the numbers to the equation for critical angle, θ c :
Remember that the angles in all these situations are defined with respect to the normal to the surface, not with respect to the plane of the surface.
Suppose a laser is placed above the surface of a smooth, freshwater pool that is of uniform depth everywhere, and aimed downwards so the light ray strikes the surface at an angle of 28° relative to the plane of the surface. At what angle, relative to the plane of the pool bottom, will the light beam strike the bottom?
This situation is illustrated in Fig. 10-6. The angle of incidence, θ, is equal to 90° minus 28°, the angle at which the laser enters the water relative to the surface. That means θ = 62°. We know that r, the index of refraction of the air, is 1.00, and also that s, the index of refraction of the water, is 1.33. We can therefore solve for the angle ø , relative to the normal N to the surface, at which the ray travels under the water:
We’re justified to go to two significant figures here, because that is the extent of the accuracy of the angular data we’re given. The angle at which the laser travels under the water, relative to the water surface, is 90° – 42°, or 48°.
Because the pool has a uniform depth, the bottom is parallel to the water surface. Therefore, by invoking the geometric rule for alternate interior angles, we can conclude that the light beam strikes the pool bottom at an angle of 48° with respect to the plane of the bottom.
Practice problems for these concepts can be found at: Reflection and Refracion Practice Test
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