Snells Law and the Refraction of Light

[spoonthrower]

The drawing shows a rectangular block of glass (n = 1.52) surrounded by liquid carbon disulfide (n = 1.63). A ray of light is incident on the glass at point A with a theta = 23.0° angle of incidence. At what angle of refraction does the ray leave the glass at point B?

http://www.boomspeed.com/boogiel80/refraction.gif

I know I have to use snells law n1theta1=n2theta2. I also know I have to use it twice, once for entering the glass and then again for when the light ray exits the glass. However, as the drawing shows, the ray that exits will not be parellel to the original ray, so I know I need to do some geometry to figure out the angles, but I have no idea where to start with the geometry. Please Help. Thanks.

 

[Doc Al]

Draw the refracted ray going from side A to side B. Consider that refracted ray to be the hypotenuse of a right triangle whose sides are the normals to sides A and B. Use that triangle to relate the angle of refraction at side A to the angle of incidence at side B.

 

[kyle8921]

Hello,

Thank you for your inquiry about Snell's Law and the refraction of light. I am happy to provide you with a response.

First, let's review Snell's Law. It 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 mediums. In this case, we have a rectangular block of glass with a refractive index of 1.52 and liquid carbon disulfide with a refractive index of 1.63.

Now, let's look at the diagram provided. As you correctly stated, we will need to use Snell's Law twice, once for entering the glass and once for exiting the glass. Let's start with entering the glass at point A.

Since we know the angle of incidence (23.0°) and the refractive index of glass (1.52), we can use Snell's Law to solve for the angle of refraction (theta2).

n1sin(theta1) = n2sin(theta2)
1.52sin(23.0°) = 1.63sin(theta2)
0.386 = 1.63sin(theta2)
sin(theta2) = 0.386/1.63 = 0.237
theta2 = sin^-1(0.237) = 13.7°

Therefore, the angle of refraction at point A is 13.7°.

Now, let's move on to exiting the glass at point B. We know that the angle of incidence (23.0°) is the same as the angle of refraction at point A (since the ray of light is entering at a perpendicular angle to the surface of the glass). Therefore, the angle of incidence at point B is also 13.7°. Since we know the angle of incidence and the refractive index of liquid carbon disulfide (1.63), we can use Snell's Law again to solve for the angle of refraction (theta3).

n1sin(theta1) = n2sin(theta3)
1.63sin(13.7°) = 1sin(theta3)
0.236 = 1sin(theta3)
sin(theta3) = 0.236/1 = 0.236
theta3 = sin^-1(0.236) = 13.