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Angles in Degrees incident angle (i) refractive angle (r) 5 10 15 20 25 30 35 40 45 50 4.1663 8.3228 12.4594 16.5650 20.6273 24.6322 28.5629 32.3993 36.1167 39.6847 2.5

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Angles in Degrees incident angle (i) refractive angle (r) 5 10 15 20 25 30 35 40 45 50 4.1663 8.3228 12.4594 16.5650 20.6273 24.6322 28.5629 32.3993 36.1167 39.6847 2.5 help-example-1
User WB Lee
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1 Answer

1 vote

The index of refraction of the substance is approximately 1.079.

The index of refraction can be determined using Snell's law, which relates the angle of incidence and the angle of refraction to the refractive indices of the two media involved.

Snell's law is given as:

n1 sin(i) = n2 sin(r)

In this case, the angle of incidence (i) is given as 45 degrees and the angle of refraction (r) is given as 40.3 degrees.

We need to determine the index of refraction (n2) of the unknown substance.

Using the formula with the given values, we can solve for n2:

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

Substituting the values, we have:

n2 = 1 sin(45) / sin(40.3)

n2 = 1 * 0.707 / 0.656

n2 = 1.079

Therefore, the index of refraction of the substance is approximately 1.079.

The probable question may be:

Angle in degrees

incident angle(i) Refractive angle(r)

5 4.1663

10 8.3228

15 12.4594

20 16.5650

25 20.6273

30 24.6322

35 28.5629

40 32.3993

45 36.1167

50 39.6847

Determine the index of refraction

User Eryc
by
8.0k points