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Chapter 23: Problem 93

A glass block \((n=1.7)\) is submerged in an unknown liquid. A ray of light inside the block undergoes total internal reflection. What can you conclude concerning the index of refraction of the liquid?

Short Answer

Expert verified
Answer: The index of refraction of the unknown liquid must be less than 1.7.

Step by step solution

01

Understand Total Internal Reflection

Total internal reflection occurs when a ray of light passes from a denser medium (higher refractive index) to a less dense medium (lower refractive index) and the angle of incidence is greater than the critical angle. This results in the complete reflection of the light back into the denser medium.
02

Write down the condition for Total Internal Reflection

For total internal reflection to occur, the incident angle must be greater than the critical angle. The critical angle is given by Snell's law: \[\sin(\theta_c) = \frac{n_2}{n_1}\] Where \(\theta_c\) is the critical angle, \(n_1\) is the index of refraction of the denser medium (glass block in our case), and \(n_2\) is the index of refraction of the less dense medium (the unknown liquid).
03

Evaluate the condition for the unknown liquid's refractive index

Since the light undergoes total internal reflection, we can conclude that the index of refraction of the liquid must be less than that of the glass block. Mathematically, this means: \[n_2 < n_1\] With the given information, we know that the index of refraction of the glass block is 1.7, so the condition becomes: \[n_2 < 1.7\]
04

Conclude the refractive index of the unknown liquid

From the analysis, we can conclude that the index of refraction of the unknown liquid must be less than 1.7 for total internal reflection to occur within the glass block.

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Most popular questions from this chapter

The angle of deviation through a triangular prism is defined as the angle between the incident ray and the emerging ray (angle \(\delta\) ). It can be shown that when the angle of incidence \(i\) is equal to the angle of refraction \(r^{\prime}\) for the emerging ray, the angle of deviation is at a minimum. Show that the minimum deviation angle \(\left(\delta_{\min }=D\right)\) is related to the prism angle \(A\) and the index of refraction \(n,\) by $$ n=\frac{\sin \frac{1}{2}(A+D)}{\sin \frac{1}{2} A} $$ [Hint: For an isosceles triangular prism, the minimum angle of deviation occurs when the ray inside the prism is parallel to the base, as shown in the figure.]
Light rays from the Sun, which is at an angle of \(35^{\circ}\) above the western horizon, strike the still surface of a pond. (a) What is the angle of incidence of the Sun's rays on the pond? (b) What is the angle of reflection of the rays that leave the pond surface? (c) In what direction and at what angle from the pond surface are the reflected rays traveling?
(a) Sunlight reflected from the still surface of a lake is totally polarized when the incident light is at what angle with respect to the horizontal? (b) In what direction is the reflected light polarized? (c) Is any light incident at this angle transmitted into the water? If so, at what angle below the horizontal does the transmitted light travel?
(a) Sunlight reflected from the smooth ice surface of a frozen lake is totally polarized when the incident light is at what angle with respect to the horizontal? (b) In what direction is the reflected light polarized? (c) Is any light incident at this angle transmitted into the ice? If so, at what angle below the horizontal does the transmitted light travel?
A beam of light in air is incident on a stack of four flat transparent materials with indices of refraction 1.20 \(1.40,1.32,\) and \(1.28 .\) If the angle of incidence for the beam on the first of the four materials is \(60.0^{\circ},\) what angle does the beam make with the normal when it emerges into the air after passing through the entire stack?
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