I don't understand snell's law ( nsin(0)=nsin(0) ) and how to use it with refraction.

also how do i apply that to this problem:
'a light ray enters a square piece of plastci (n=1.39) near a corner. as it encounters the adjacent side it gets refracted along the surface (it hits at the critical angle). what was the original angle with which the ray hit the plastic from air?'
i know the answer (75 degrees) but i don't know how to get there.
please help!

What does "near a corner" mean? Does that mean the ray is internally reflected once?

i have no idea, i just wrote what the question said. but yes that's what i interpreted it to mean

Then start at the end of the ray.

1.39*sinThetainside=1*sin90
theta inside =46 deg.
Now take that angle back to where it was reflected, so it incident angle at the reflection was 46, then trace back to the corner side, so it leaves the surface refracted at 54 deg, no use Snell's law to find the ray angle outside ..

1.39sin54=1*sinTheta
and I think you will get your 75deg. Draw the figure to make certain you understand it.

oops, typo it is 44 deg, not 54 deg. The sum has to be 90 between 46 and 44.

Snell's law is a fundamental principle in optics that relates the angles and indices of refraction of light as it passes through different mediums. It can be stated as follows: n₁sin(θ₁) = n₂sin(θ₂), where n₁ and n₂ are the indices of refraction of the two media, and θ₁ and θ₂ are the angles of incidence and refraction, respectively.

In the context of refraction, Snell's law tells us how light bends when it passes through a boundary between two different media. The law states that the ratio of the sine of the angle of incidence in the first medium to the sine of the angle of refraction in the second medium is equal to the ratio of the indices of refraction of the two media.

Now, let's apply Snell's law to the given problem of the light ray entering a square piece of plastic. We know that n₁ (the index of refraction of air) is approximately 1.00, and n₂ (the index of refraction of plastic) is given as 1.39.

The problem states that the light ray hits the plastic at the critical angle. The critical angle is the angle of incidence at which the refracted ray travels along the boundary surface, rather than entering the second medium.

To find the original angle of incidence, we need to determine the angle at which the refracted ray travels along the surface. At this angle, the angle of refraction, θ₂, is 90 degrees.

Using Snell's law, we have:

n₁sin(θ₁) = n₂sin(θ₂)

Since sin(90) = 1, we can substitute the given values:

1.00sin(θ₁) = 1.39sin(90)

Simplifying, we have:

sin(θ₁) = 1.39

Now, we need to determine the angle θ₁ that satisfies sin(θ₁) = 1.39. By taking the inverse sine or arcsine of 1.39, we can find the angle.

Using a calculator, arcsin(1.39) is approximately 75 degrees.

Therefore, the original angle with which the light ray hit the plastic from air is approximately 75 degrees.