Solve the following problems using Snell\'s law. The first problem is doue for y

Question

Solve the following problems using Snell's law. The first problem is doue for you as an example. 1. Air has an index of refraction equal to L.0 and glass has an index of refraction equal to 1.5. Light travel from air into glass with an angle of incidence ,- 25°. What is the refractive angle 8,? We can solve fr q from the relation: n, sing-pang sine, , sino,= 1.5 sin25°-1.5 (0,4230.282 and the angle e, is given by the inverse sine of 0.285: sin-1(0.282) = 16.3s-e 2. Light travels from glass into air. Theincidentangleq-30·What is the angle of refraction? 3. Light travels from air into glass. The angle of refraction is e-30. what is the ingle 4. Light travels from air into diamond. The index of refraction for diamond is 2.4 and the incident angle 0,-30°. What is the angle of refraction? If the incident and refraction angles are 30 degrees and 45 degrees, respectively, which material has the larger index of refraction and what is the ratio of the refraction indices? 5. 6. Air has an index of refraction equal to 1.0 and glass has an index of refraction equal to 1.5. Light travels from glass into air. Caleulate the incident angle for which the refractive angle equals 90 degrees. When the angle of refraction becomes 90 degrees we have a very special and interesting situation. The incident angle that corresponds to this case is called the critical angle. When the incident angle becomes greater that the critical angle the incident light will now be reflected at the surface. This is called total internal reflection. Caleulate the critical angle (the angle of incidence) as we did on problem 6 for light travelling from diamond to air. 7. 8. Calculate the critical angle of refraction for the water-air interface. The index of refraction for water is 1.3

Explanation / Answer

2)

theta1 = 30 degree

n1 = 1

n2 = 1.5

Using snell's law
n1 * sin(i) = n2 * sin(r)

1 * sin(30) = 1.5 * sin(r)

r = arsin(0.333)

the angle of refraction is arsin(0.333)

3)

r = 30 degree

n1 = 1

n2 = 1.5

Using snell's law
n1 * sin(i) = n2 * sin(r)

1 * sin(i) = 1.5 * sin(30)

i = arsin(0.75)

the angle of incidence is arsin(0.75)

4)

i = 30 degree

n1 = 1

n2 = 2.4

Using snell's law

n1 * sin(i) = n2 * sin(r)

1 * sin(30) = 2.4 * sin(r)

r = arsin(0.2083)

the angle of refraction is arsin(0.2083)

5)

i =30 degree

r = 45 degree

Using snell's law

n1 * sin(i) = n2 * sin(r)

n1/n2 = sin(45)/sin(30)

n1/n2 = 1.414

the ratio of refractive indices is 1.414

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