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Chapter 33: Problem 50

Light is incident normally on the short face of a \(30^{\circ}-60^{\circ}-90^{\circ}\) prism (Fig. \(\mathbf{P 3 3 . 5 0}\) ). A drop of liquid is placed on the hypotenuse of the prism. If the index of refraction of the prism is \(1.56,\) find the maximum index that the liquid may have for the light to be totally reflected.

Short Answer

Expert verified
The maximum index of refraction that the liquid may have for the light to be totally reflected is 3.12

Step by step solution

01

Understand conditions for total internal Reflection

For total internal reflection to occur, the light must be traveling from a medium of higher refractive index \(n_1\) to a medium of lower refractive index \(n_2\) and the angle of incidence must be greater than the so-called critical angle. In this case, the light is traveling from the prism to the liquid. Therefore, the refractive index of the prism must be higher than that of the liquid.
02

Apply Snell's Law

Snell's law states that the ratio of the sine of the angles of incidence and refraction is equivalent to the ratio of indices of refraction. Using the terms from the 30-60-90 prism: \(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\), where \(\theta_1 = 90^{\circ}\), \(n_1 = 1.56\) (index of refraction of the prism) and \(\theta_2 = 30^{\circ}\), as the ray is incident on the 60 degree angle, so it refracts at an angle of 30 degrees inside the prism. Since we know \(sin(90^{\circ}) = 1\) and \(sin(30^{\circ}) = 0.5\), we can find \(n_2\), the refractive index of the liquid.
03

Solve for the refractive index of the liquid

Rearrange the equation from Step 2 to solve for \(n_2\): \(n_2 = \frac{n_1 \sin(\theta_1)}{\sin(\theta_2)}\). Substitute \(n_1 = 1.56\), \(\sin(\theta_1) = 1\) and \(\sin(\theta_2) = 0.5\) into the equation: \(n_2 = \frac{1.56 \times 1}{0.5}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Refractive Index
The refractive index, denoted as n, is a measure of how much a beam of light bends, or refracts, as it enters a material from a vacuum or from air, which we consider as the standard medium with a refractive index of 1. This bending occurs because light travels at different speeds in different media. When light enters a material with a higher refractive index, it slows down and bends towards the normal 鈥� an imaginary line perpendicular to the surface at the point of incidence.

To visualize this, imagine you're pushing a shopping cart from a smooth floor onto a carpet: the wheels hitting the carpet first slow down, causing the cart to turn towards the carpet. Similarly, light changes direction when entering a new medium, though the principles governing the phenomenon are different.

In our exercise, the prism has a refractive index of 1.56, which means light moves slower in the prism than in air. The higher the refractive index of a medium, the more significantly light will bend upon entry.
Snell's Law
One of the fundamental principles in optics is Snell's Law, which relates the angles of incidence and refraction to the refractive indices of the two media involved. The law is mathematically expressed as:
\[\begin{equation} n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \end{equation}\]
where
  • n_1 and n_2 are the refractive indices of the first and second media, respectively,
  • \theta_1 is the angle of incidence, and
  • \theta_2 is the angle of refraction.

Why is it practical?

Snell's Law allows us to calculate how much light will bend when transitioning between materials. In our case, we use it to find the conditions under which total internal reflection will occur, namely, the maximum refractive index the liquid can have for light to still reflect entirely inside the prism.

Critical Angle
A pivotal concept in understanding total internal reflection is the critical angle. This is the angle of incidence above which light doesn't exit a medium but instead reflects back into it. It occurs when light moves from a medium with a higher refractive index to one with a lower refractive index.

The critical angle can be calculated using the formula:
\[\begin{equation} \theta_c = \sin^{-1}(\frac{n_2}{n_1}) \end{equation}\]
where \(\theta_c\) is the critical angle, n_1 is the refractive index of the denser medium, and n_2 is the refractive index of the less dense medium. When applying this to the exercise, we recognize that the refractive index of the liquid must be lower than the prism for total internal reflection to be possible, and hence, can calculate the maximum index the liquid may have.

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Most popular questions from this chapter

A layer of liquid sits on top of the horizontal surface of a transparent solid. For a ray traveling in the solid and incident on the interface of the two materials, the critical angle is \(38.7^{\circ}\). (a) For a ray traveling in the solid and reflecting at the interface with the liquid, for what incident angle with respect to the normal is the reflected ray \(100 \%\) polarized? (b) What is the polarizing angle if the ray is traveling in the liquid?

Unpolarized light of intensity \(20.0 \mathrm{~W} / \mathrm{cm}^{2}\) is incident on two polarizing filters. The axis of the first filter is at an angle of \(25.0^{\circ}\) counterclockwise from the vertical (viewed in the direction the light is traveling), and the axis of the second filter is at \(62.0^{\circ}\) counterclockwise from the vertical. What is the intensity of the light after it has passed through the second polarizer?

Given \(\quad\) small samples of three liquids, you are asked to determine their refractive indexes. However, you do not have enough of each liquid to measure the angle of refraction for light refracting from air into the liquid. Instead, for each liquid, you take a rectangular block of glass \((n=1.52)\) and place a drop of the liquid on the top surface of the block. You shine a laser beam with wavelength \(638 \mathrm{nm}\) in vacuum at one side of the block and measure the largest angle of incidence \(\theta_{a}\) for which there is total internal reflection at the interface between the glass and the liquid (Fig. \(\mathbf{P 3 3 . 5 6}\) ). Your results are given in the table: $$ \begin{array}{l|lll} \text { Liquid } & A & B & C \\ \hline \boldsymbol{\theta}_{\boldsymbol{a}}\left({ }^{\circ}\right) & 52.0 & 44.3 & 36.3 \end{array} $$ What is the refractive index of each liquid at this wavelength?

(a) A tank containing methanol has walls \(2.50 \mathrm{~cm}\) thick made of glass of refractive index \(1.550 .\) Light from the outside air strikes the glass at a \(41.3^{\circ}\) angle with the normal to the glass. Find the angle the light makes with the normal in the methanol. (b) The tank is emptied and refilled with an unknown liquid. If light incident at the same angle as in part (a) enters the liquid in the tank at an angle of \(20.2^{\circ}\) from the normal, what is the refractive index of the unknown liquid?

A horizontal, parallel-sided plate of glass having a refractive index of 1.52 is in contact with the surface of water in a tank. A ray coming from above in air makes an angle of incidence of \(35.0^{\circ}\) with the normal to the top surface of the glass. (a) What angle does the ray refracted into the water make with the normal to the surface? (b) What is the dependence of this angle on the refractive index of the glass?
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