Question

A flat piece of glass covers the top of a vertical cylinder that is completely filled with water. If a ray of light traveling in the glass is incident on the interface with the water at an angle of 0a= 36.0 degrees , the ray refracted into the water makes an angle of 49.6 degrees with the normal to the interface.What is the smallest value of the incident angle 0a for which none of the ray refracts into the water? Express your answer with the appropriate units.

Answer 1

`\theta = 50.5 degree`

Step-by-step explanation:

As we know that the refractive index of the water is

`\mu_w = \mu_1`

also we know that refractive index of the glass is

`\mu_g = \mu_2`

now we know by Snell's law

`\mu_1 sin\theta_1 = \mu_2 sin\theta_2`

so we have

`\mu_1 sin36 = \mu_2 sin49.6`

`(\mu_1)/(\mu_2) = 1.2956`

now if no light will refract into the water then in that case

`\mu_1 sin\theta = \mu_2 sin90`

now we have

`(1.2956) sin\theta = 1`

`sin\theta = 0.77`

`\theta = 50.52 degree`