When the rectangular metal tank in the Figure is filled to the top with an unknown liquid, the observer O, with eyes level with the top of the tank, can just see the corner. A ray that refracts towardOthe top surface of the liquid is shown. IfD=85.0cmsoL=1.10m, what is the index of refraction of the liquid?

Figure:

Short Answer

Expert verified

The refractive index of the liquid isn1=1.26.

Step by step solution

01

Given

Consider the given parameters shown below:

L=1.10mD=85cm=0.85mθ2=90°

  1. The Refractive index of vacuum is n2=1.
02

Determine the concept

Use the formula of Snell’s law, which gives the relation between the angle of reflection and the angle of refraction. Before that, the angle of incidence using the equation of tangent ratio can be calculated

03

Determine the refractive index of liquid

Expression for an angle of incidence is as follows:

tanθ1=LDtanθ1=1.100.85θ1=52.31°

Now use Snell’s law:

n1sinθ1=n2sinθ2n1sin52.31=1sin90n1=1.26

So the refractive index of the liquid isn1=1.26 .

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Most popular questions from this chapter

Question: In Fig. 33-64, a light ray in air is incident on a flat layer of material 2 that has an index of refraction n2=1.5 . Beneath material 2is material 3 with an index of refraction n3.The ray is incident on the air–material 2 interface at the Brewster angle for that interface. The ray of light refracted into material 3happens to be incident on the material 2-material 3 interface at the Brewster angle for that interface. What is the value of n3 ?

Suppose we rotate the second sheet in Fig. 33-15a, starting with the polarization direction aligned with the y axis(θ=00)and ending with it aligned with the x-axis(θ=90°). Which of the four curves in Fig. 33-26 best shows the intensity of the light through the three-sheet system during this90°rotation?

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