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Chapter 26: Problem 27

Concept Simulation 26.1 at illustrates the concepts that are pertinent to this problem. A ray of light is traveling in glass and strikes a glass-liquid interface. The angle of incidence is \(58.0^{\circ},\) and the index of refraction of glass is \(n=1.50 .\) (a) What must be the index of refraction of the liquid such that the direction of the light entering the liquid is not changed? (b) What is the largest index of refraction that the liquid can have, such that none of the light is transmitted into the liquid and all of it is reflected back into the glass?

Short Answer

Expert verified
(a) The index must be 1.50. (b) The largest index is 1.272.

Step by step solution

01

Understanding Reflection and Refraction

Light will travel without refraction (direction change) at the interface if the angle of refraction is the same as the angle of incidence. For zero refraction, the index of refraction of the liquid must be equal to the index of refraction of the glass.
02

Calculating the Index of Refraction for Zero Refraction

According to Snell's Law, \(n_{1} \sin \theta_{1} = n_{2} \sin \theta_{2}\). If the direction doesn't change, \(\theta_{1} = \theta_{2}\), hence \(n_{1} = n_{2}\). Thus, for no change in direction, the liquid's index of refraction \(n = 1.50\).
03

Reflection via Total Internal Reflection

For complete reflection, none of the light enters the liquid, which implies total internal reflection. This occurs when the angle of incidence is greater than or equal to the critical angle \(\theta_c\).
04

Calculating the Critical Angle

The critical angle \(\theta_c\) can be found using \(\sin \theta_c = \frac{n_{2}}{n_{1}}\), where \(n_{1} = 1.50\) and \(n_{2}\) is the unknown refractive index of the liquid. Set \(\theta_c = 58^{\circ}\) and solve for \(n_{2}\):\[\sin 58^{\circ} = \frac{n_{2}}{1.50}\]Solving for \(n_{2}\) gives \(n_{2} = 1.50 \sin 58^{\circ}\).
05

Finding the Largest Index for Total Internal Reflection

Calculate \(n_{2}\) with \(\theta_c = 58^{\circ}\):\[n_{2} = 1.50 \cdot 0.848\]\[n_{2} = 1.272\]Thus, the largest index of refraction the liquid can have for total internal reflection to occur is approximately 1.272.

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Key Concepts

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

Total Internal Reflection
Total internal reflection is a fascinating optical phenomenon. It occurs when a light ray traveling through a medium hits a boundary at an angle larger than a specific critical angle, causing it to reflect entirely within the original medium. This means no part of the light ray passes into the second medium.
For this to happen, two conditions must be met:
  • The light must be traveling from a medium with a higher index of refraction to one with a lower index.
  • The angle of incidence must be larger than the critical angle specific to the mediums involved.
When the critical angle is surpassed, total internal reflection ensures that all the light is reflected, and none is refracted. This principle is the basis for many technologies, including optical fibers.
Index of Refraction
The index of refraction, often referred to as the refractive index, is a measure of how much a substance can bend light. Technically speaking, it is a dimensionless number that describes how light propagates through a medium.
Here are key points on the index of refraction:
  • If the refractive index of a medium is high, light travels slower through it than in a medium with a lower refractive index.
  • The equation that governs this behavior is called Snell's Law, expressed as: \[ n_{1} \sin \theta_{1} = n_{2} \sin \theta_{2} \]where \( n \) represents the refractive index, and \( \theta \) is the angle of the light ray.
This concept is crucial in understanding how light behaves at interfaces, such as between glass and liquid. By measuring an angle of incidence and knowing the indices of refraction of the interacting substances, the resulting path of light can be predicted.
Critical Angle
The critical angle is specific to a pair of media, such as glass and liquid. It's the angle of incidence above which total internal reflection occurs. Why does this happen? When light travels from a denser medium to a less dense one, refraction directs the light away from the normal line at the boundary. If you decrease the density enough or increase the angle, the light can't refract out; it reflects inside instead.To calculate the critical angle, you use the formula:\[\sin \theta_c = \frac{n_{2}}{n_{1}}\]
  • \( n_{1} \) and \( n_{2} \) are the indices of refraction for the initial and second medium, respectively.
  • If the angle of incidence is greater than \( \theta_c \), all light reflects back into the denser medium (total internal reflection).
Knowing the critical angle helps determine scenarios where total internal reflection can occur, which is vital for applications like designing optical devices that rely on light being guided without loss.

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Most popular questions from this chapter

The back wall of a home aquarium is a mirror that is a distance \(L\) away from the front wall. The walls of the tank are negligibly thin. A fish, swimming midway between the front and back walls, is being viewed by a person looking through the front wall. (a) Does the fish appear to be at a distance greater than, less than, or equal to \(\frac{1}{2} L\) from the front wall? (b) The mirror forms an image of the fish. How far from the front wall is this image located? Express your answer in terms of \(L\). (c) Assume that your answer to Question (b) is a distance \(D\). Does the image of the fish appear to be at a distance greater than, less than, or equal to \(D\) from the front wall? (d) Could the image of the fish appear to be in front of the mirror if the index of refraction of water were different than it actually is, and, if so, would the index of refraction have to be greater than or less than its actual value? Explain each of your answers. The distance between the back and front walls of the aquarium is \(40.0 \mathrm{~cm} .\) (a) Calculate the apparent distance between the fish and the front wall. (b) Calculate the apparent distance between the image of the fish and the front wall. Verify that your answers are consistent with your answers to the Concept Questions.

Sunlight strikes a diamond surface. At what angle of incidence is the reflected light completely polarized?

A scuba diver, submerged under water, looks up and sees sunlight at an angle of \(28.0^{\circ}\) from the vertical. At what angle, measured from the vertical, does this sunlight strike the surface of the water?

Consult Interactive Solution 26.51 at to review the concepts on which this problem depends. A camera is supplied with two interchangeable lenses, whose focal lengths are 35.0 and \(150.0 \mathrm{~mm}\). A woman whose height is \(1.60 \mathrm{~m}\) stands \(9.00 \mathrm{~m}\) in front of the camera. What is the height (including sign) of her image on the film, as produced by (a) the 35.0 -mm lens and (b) the 150.0 -mm lens?

At offers a review of the concepts that play roles in this problem. A diverging lens \((f=-10.0 \mathrm{~cm})\) is located \(20.0 \mathrm{~cm}\) to the left of a converging lens \((f=30.0 \mathrm{~cm})\). A \(3.00-\mathrm{cm}\) -tall object stands to the left of the diverging lens, exactly at its focal point. (a) Determine the distance of the final image relative to the converging lens. (b) What is the height of the final image (including the proper algebraic sign)?
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