Question

The refractive index of a transparent material can be determined by measuring the critical angle when the solid is in air. If θ_c= 41.5° what is the index of refraction of the material?

Tries 0/12

A light ray strikes this material (from air) at an angle of 37.1° with respect to the normal of the surface. Calculate the angle of the reflected ray (in degrees).

Tries 0/12

Calculate the angle of the refracted ray (in degrees).

Tries 0/12

Assume now that the light ray exits the material. It strikes the material-air boundary at an angle of 37.1° with respect to the normal. What is the angle of the refracted ray?

Tries 0/12

Answer 1

The refractive index of a transparent material can be determined by measuring the critical angle when the solid is in air. If θc= 41.5° what is the index of refraction of the material?

    Tries 0/12     
A light ray strikes this material (from air) at an angle of 37.1° with respect to the normal of the surface. Calculate the angle of the reflected ray (in degrees).

    Tries 0/12     
Calculate the angle of the refracted ray (in degrees).

    Tries 0/12     
Assume now that the light ray exits the material. It strikes the material-air boundary at an angle of 37.1° with respect to the normal. What is the angle of the refracted ray?
    Tries 0/12     

Here ,

critical angle , thetac = 41.5 degree

Using snell's law

n* sin(thetaC) = sin(90) * 1

n * sin(41.5) = 1

n = 1.51

the refractive index of the material is 1.51
---------------------------------------

as angle of reflection = anfle of incideance

angle of reflection = 37.1 degre

-------------------------------

let the angle of refraction is r

Using snell's law

1.51 * sin(r) = 1 * sin(37.1)

solving for r

r = 23.52 degree

the angle of refraction is 23.52 degree

----------------------------------
Now,

Using snell's law

1.51 * sin(37.1) = 1 * sin(r)

solving for r

r = 65.6 degree

the angle of refraction is 65.6 degree