Skip to main content
Ch 33: The Nature and Propagation of Light
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 33: The Nature and Propagation of LightProblem 17a
Chapter 33, Problem 17a

The critical angle for total internal reflection at a liquid–air interface is 42.5°. If a ray of light traveling in the liquid has an angle of incidence at the interface of 35.0°, what angle does the refracted ray in the air make with the normal?

Verified step by step guidance
1
Understand the concept of total internal reflection: It occurs when a ray of light travels from a medium with a higher refractive index to a medium with a lower refractive index, and the angle of incidence exceeds the critical angle, resulting in no refraction and complete reflection within the medium.
Identify the given values: The critical angle is 42.5°, and the angle of incidence is 35.0°. Since the angle of incidence is less than the critical angle, refraction will occur.
Apply Snell's Law to find the angle of refraction. Snell's Law is given by: n1⁢sin⁢θ1=n2⁢sin⁢θ2, where n1 and n2 are the refractive indices of the liquid and air respectively, and θ1 and θ2 are the angles of incidence and refraction.
Assume the refractive index of air is approximately 1. Use the critical angle to find the refractive index of the liquid: n1=1sin⁢42.5°. Calculate n1.
Substitute the values into Snell's Law to solve for the angle of refraction θ2: sin⁢θ2=n1⁢sin⁢35.0°. Calculate θ2.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
6m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Total Internal Reflection

Total internal reflection occurs when a light ray traveling in a denser medium hits the boundary with a less dense medium at an angle greater than the critical angle, resulting in the light being completely reflected back into the denser medium. The critical angle is the minimum angle of incidence at which this phenomenon occurs.
Recommended video:
Guided course
05:29
Total Internal Reflection

Snell's Law

Snell's Law describes how light bends when it passes from one medium to another. It states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant and equal to the ratio of the indices of refraction of the two media. This law is essential for calculating the angle of refraction when light crosses an interface.
Recommended video:
Guided course
07:59
Snell's Law

Critical Angle

The critical angle is the angle of incidence above which total internal reflection occurs for light traveling from a denser to a less dense medium. It is determined by the refractive indices of the two media and is calculated using Snell's Law. For angles less than the critical angle, light will refract into the second medium.
Recommended video:
Guided course
01:04
Critical Angle
Related Practice
Textbook Question

It is proposed to store 1.00 kWh = 3.60 × 106 J of electrical energy in a uniform magnetic field with magnitude 0.600 T. If instead this amount of energy is to be stored in a volume (in vacuum) equivalent to a cube 40.0 cm on a side, what magnetic field is required?

2101
views
Textbook Question

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° with the normal to the top surface of the glass. What angle does the ray refracted into the water make with the normal to the surface?

1906
views
Textbook Question

Light enters a solid pipe made of plastic having an index of refraction of 1.60. The light travels parallel to the upper part of the pipe (Fig. E33.15). You want to cut the face AB so that all the light will reflect back into the pipe after it first strikes that face. If the pipe is immersed in water of refractive index 1.33, what is the largest that u can be?

2006
views
Textbook Question

The indexes of refraction for violet light (λ = 400 nm) and red light (λ= 700 nm) in diamond are 2.46 and 2.41, respectively. A ray of light traveling through air strikes the diamond surface at an angle of 53.5° to the normal. Calculate the angular separation between these two colors of light in the refracted ray.

1279
views
Textbook Question

Light enters a solid pipe made of plastic having an index of refraction of 1.60. The light travels parallel to the upper part of the pipe (Fig. E33.15). You want to cut the face AB so that all the light will reflect back into the pipe after it first strikes that face. What is the largest that u can be if the pipe is in air?

1766
views
Textbook Question

At the very end of Wagner's series of operas Ring of the Nibelung, Brünnhilde takes the golden ring from the finger of the dead Siegfried and throws it into the Rhine, where it sinks to the bottom of the river. Assuming that the ring is small enough compared to the depth of the river to be treated as a point and that the Rhine is 10.0 m deep where the ring goes in, what is the area of the largest circle at the surface of the water over which light from the ring could escape from the water?

1883
views