Question

a ray of light propagates from a material with refractive index 1.53 to a material of refractive index 1.45. what is the maximum angle of incidence the incoming ray can have if it is to be transmitted?

Answer 1

The maximum angle of incidence a ray of light can have to be transmitted from a material with a refractive index of 1.53 to a material with a refractive index of 1.45 is approximately 70.52 degrees.

Step-by-step explanation:

The maximum angle of incidence a ray of light can have in order to be transmitted from a material with a refractive index of 1.53 to a material with a refractive index of 1.45 can be calculated using Snell's law. Snell's law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the refractive indices of the two materials:
n1 * sin(angle of incidence) = n2 * sin(angle of refraction)

In this case, the incident angle is the maximum angle of incidence we want to find, so we can rearrange the equation to solve for it:

angle of incidence = arcsin((n2/n1) * sin(angle of refraction))

Substituting the given values into the equation, we get:

angle of incidence = arcsin((1.45/1.53) * sin(90)

The sine of 90 degrees is 1, so the equation simplifies to:

angle of incidence = arcsin(0.947)

Using a calculator, the angle of incidence is approximately 70.52 degrees.

Therefore, the maximum angle of incidence the incoming ray can have in order to be transmitted is 70.52 degrees.