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Chapter 35: Problem 19

Light traveling in water of refractive index \(1.33\) is incident on a plate of glass with index of refraction 1.53. At what angle of incidence is the reflected light fully polarized?

Short Answer

Expert verified
Approximately \(49.6^\text{o}\).

Step by step solution

01

Understand Brewster's Angle

When light is incident at the Brewster's angle, the reflected light is fully polarized. Brewster鈥檚 law states that the Brewster's angle occurs when the refracted and reflected rays are perpendicular to each other.
02

Recall the Brewster's Angle Formula

Brewster's angle (\theta_B) can be found using the formula: \[\tan(\theta_B) = \frac{n_2}{n_1} \] where \(n_1\) is the refractive index of the initial medium (water) and \(n_2\) is the refractive index of the second medium (glass).
03

Substitute the Known Values

Substitute the given values into the formula: \[\tan(\theta_B) = \frac{1.53}{1.33} \]
04

Solve for Brewster's Angle

Calculate \(\theta_B\): \[\theta_B = \tan^{-1}\frac{1.53}{1.33} \] Using a calculator, this evaluates to approximately \(49.6^\text{o} \).

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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 is a measure of how much light slows down when it enters a medium from another medium. When light passes from air into water or glass, it bends.
This bending happens because light travels at different speeds in different substances.
The refractive index, denoted as \(n\), is defined by the ratio of the speed of light in a vacuum to the speed of light in the medium.
Mathematically, it's expressed as: ewline












This helps minimize errors and ensures your answer is as accurate as possible.

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Most popular questions from this chapter

A converging lens with a focal length of \(+20 \mathrm{~cm}\) is located \(10 \mathrm{~cm}\) to the left of a diverging lens having a focal length of \(-15 \mathrm{~cm}\). If an object is located \(40 \mathrm{~cm}\) to the left of the converging lens, locate and describe completely the final image formed by the diverging lens.

A short straight object of length \(L\) lies along the central axis of a spherical mirror of focal length \(f\), a distance \(o\) from the mirror. (a) Show that its image in the mirror has a length \(L^{\prime}\) where $$L^{\prime}=L\left(\frac{f}{o-f}\right)^{2}$$ (Hint: Locate the two ends of the object.) (b) Show that the longitudinal magnification \(m^{\prime}\left(=L^{\prime} / L\right)\) is equal to \(m^{2}\), where \(m\) is the lateral magnification.

You may have observed people with cameras behaving strangely: (a) At a conference in North Carolina, one of the physics graduate students (who should have known better!) tried to take a picture of the overheads projected on a white screen. Since the room was darkened, he used a flash. Explain why this is a bad idea and what his pictures are likely to show. (b) Someone mentioned to the student that he probably should not be using his flash, so he turned it off. He then proceeded to try and take pictures of the participants in the darkened room! Explain why this is a bad idea and what his pictures are likely to show. (c) A woman on an airplane at night with a camera was impressed with the view of the city lights in the dark as the plane flew over Washington, D.C. She stood back in the aisle with her camera and tried to take a picture through the window using her flash. Explain why this is a bad idea and what her pictures are likely to show.

A concave shaving mirror has a radius of curvature of \(35.0 \mathrm{~cm} .\) It is positioned so that the (upright) image of a man's face is \(2.50\) times the size of the face. How far is the mirror from the face?

Two thin lenses of focal lengths \(f_{1}\) and \(f_{2}\) are in contact. Show that they are equivalent to a single thin lens with $$f=\frac{f_{1} f_{2}}{f_{1}+f_{2}}$$ as its focal length.
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