Question

A layer of liquid B floats on liquid A. A ray of light begins in liquid A and undergoes total internal reflection at the interface between the liquids when the angle of incidence exceeds 38.0°. When liquid B is replaced with liquid C, total internal reflection occurs for angles of incidence greater than 49.5°. Find the ratio nB/nC of the refractive indices of liquids B and C. nB/nC =?

Answer 1

To find the ratio of refractive indices of liquids B and C (nB/nC), we can use the concept of total internal reflection and Snell's law. By determining the critical angles for each liquid and dividing nB by nC, we can calculate the desired ratio.

Step-by-step explanation:

To find the ratio of refractive indices of liquids B and C (nB/nC), we can start by using the concept of total internal reflection. Total internal reflection occurs when the incident angle is greater than the critical angle. The critical angle can be determined by using Snell's law: sin(critical angle) = n2/n1.

From the given information, we know that the incident angle for total internal reflection with liquid B is 38.0°, and with liquid C is 49.5°. We can use these values to determine the critical angles for each liquid. Then, by dividing the refractive index of liquid B (nB) by the refractive index of liquid C (nC), we can find the desired ratio nB/nC.

To calculate the critical angles:

1. For liquid B, sin(critical angle of B) = n1B / n1A, where n1B is the refractive index of liquid B and n1A is the refractive index of liquid A (which is not given).
2. For liquid C, sin(critical angle of C) = n1C / n1A, where n1C is the refractive index of liquid C and n1A is the refractive index of liquid A.

Once we know the critical angles, we can determine the desired ratio by dividing nB by nC: nB/nC = sin(critical angle of C) / sin(critical angle of B).

By solving these equations, we can determine the value of nB/nC.

Answer 2

The ratio of the refractive indices of the liquids is 1.22

Solution:

Critical angle for A and B interface,
`\theta_(AB) = 38.0^(\circ)`

Critical angle for A and B interface,
`\theta_(AC) = 49.5^(\circ)`

Now,

From the relation in between the critical angle and refractive index:

`n = (1)/(\theta_(c))` (1)

where

n = rafractive index

`\theta_(c)` = critical angle

Thus

For AB interface:

`n_(B) = (1)/(sin\theta_(AB)) = (1)/(sin(38.0^(\circ))) = 1.606`

For AC interface:

`n_(C) = (1)/(sin\theta_(AC)) = (1)/(sin(49.5^(\circ))) = 1.315`

Thus the ratio of the refractive indices of these liquids can be given as:

`(n_(B))/(n_(C)) = (1.606)/(1.315)`

`(n_(B))/(n_(C)) = 1.22`