You have two containers half full with liquid 1. You then fill the two containers respectively with a layer of liquid 2 and 3. A ray of light travelling through the bottom layer (liquid 1) and into the top layer experiences total internal reflection off liquid 2 when the angle of incidence exceeds 44.5° and for liquid 3 total internal reflection occurs for angles of incidence greater than 50.5°. Determine the ratio of the refractive indices of liquids 2 and 3 (n2/n3).
Wont Snell's law give this to you directly?
To solve this problem, we will use the principles of Snell's Law and total internal reflection.
Let's assume that the refractive index of liquid 1 is n1.
When the ray of light passes from liquid 1 into liquid 2, we can apply Snell's Law:
n1 * sin(angle of incidence 1) = n2 * sin(angle of refraction 1)
Since we know that total internal reflection occurs when the angle of incidence exceeds 44.5° in liquid 2, we can use this value to find the critical angle for liquid 2. The critical angle (angle of incidence) is the angle where the angle of refraction becomes 90°.
So, sin(critical angle 2) = n2 / n1
sin(critical angle 2) = n2 / 1 (as we assume n1 = 1 for simplicity)
Now, when the ray of light passes from liquid 2 into liquid 3, we can apply Snell's Law again:
n2 * sin(angle of incidence 2) = n3 * sin(angle of refraction 2)
We are given that total internal reflection occurs for angles of incidence greater than 50.5° in liquid 3. So, we can use this value to find the critical angle for liquid 3:
sin(critical angle 3) = n3 / n2
Now, we can solve for the ratio of n2 and n3:
n2 / n3 = sin(critical angle 2) / sin(critical angle 3)
Therefore, to determine the ratio of the refractive indices of liquids 2 and 3 (n2/n3), you need to calculate the sine of the critical angles of liquid 2 and liquid 3, and then divide the calculated value of sin(critical angle 2) by sin(critical angle 3).