An optic fiber is made of clear plastic with index of refraction of 1.50. For what angles of incidence will light remain within the plastic?

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Let crictical angle be `theta_c` ,so that light remain in plastic

We know

`sin(theta_c)=eta_2/eta_1`

where `eta_2 =1.0`refreactive index of the air , and

`eta_1=1.5` ,refractive index of the plastic.

Thus

`sin(theta_c)=1/1.5=.6666`

`theta_c=41.8^o`

For angles greater that 41.8o with respect to the normal .

Or angles smaller than (90o - 41.8o) = 48.2o with respect to the surface .

When a beam of light passes from one medium to another the beam bends if the two mediums do not have the same refractive index.

If the angle of incidence is A and the angle of refraction is B, the two are related by `(sin A)/(sin B) = (n2)/(n1)` where n1 is the refractive index of the first medium and n2 is the refractive index of the second.

The optic fiber is made of a material with refractive index 1.5, the refractive index of air is 1. A beam of light undergoes total internal reflection when the angle of incidence is equal to `sin^-1((n2)/(n1))` . For n2 = 1 and n1 = 1.5, `sin^-1(1/1.5) ~~ 41.81` degrees.

If the angle of incidence is equal to 41.81 or greater the beam of light does not leave the optic fiber.

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