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Chapter 26: Problem 113

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?

Short Answer

Expert verified
20.7掳

Step by step solution

01

Understanding the Scenario

The scenario involves light moving from water to air. The sunlight is observed at an angle of 28.0掳 from the vertical, which is the angle of refraction. We need to find the angle of incidence, at which the sunlight strikes the water surface.
02

Apply Snell's Law

We apply Snell's Law, which quantifies how light bends when it passes from one medium into another. Snell's Law is as follows: \[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]For this scenario, \( n_1 = 1.33 \) (refractive index of water), \( n_2 = 1 \) (refractive index of air), \( \theta_2 = 28.0^{\circ} \). We need to find \( \theta_1 \).
03

Rearrange Snell's Law

Rearrange the given formula to solve for the angle of incidence \( \theta_1 \): \[ \sin(\theta_1) = \frac{n_2}{n_1} \sin(\theta_2) \]
04

Calculate the Sine of the Angle of Incidence

Substitute the known values into the equation: \[ \sin(\theta_1) = \frac{1}{1.33} \cdot \sin(28.0^{\circ}) \] Calculate \( \sin(28.0^{\circ}) \approx 0.4695 \) and perform the division to find \( \sin(\theta_1) \).
05

Solve for the Angle of Incidence

Use the inverse sine function to find \( \theta_1 \). Calculate: \[ \theta_1 = \sin^{-1}(0.3531) \approx 20.7^{\circ} \]Thus, the angle of incidence from the vertical is 20.7掳.

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Key Concepts

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

Angle of Refraction
When light transitions between different mediums, its path bends or refracts. The angle of refraction helps describe this phenomenon. It is the angle between the refracted ray (the path light takes as it enters a new medium) and a line that is perpendicular to the boundary between the two mediums.
In our case, the diver sees sunlight at a 28掳 angle. This is the angle of refraction from when the light moves from water to air.
Understanding the angle of refraction is crucial because it gives insight into how light behaves across various surfaces. This angle is dependent on the relative speeds of light in the two mediums, as described by Snell's Law.
Refractive Index
Refractive index, often denoted by the symbol \(n\), is a measure of how much the light slows down in a particular medium. It's calculated as the ratio of the speed of light in a vacuum to the speed of light in the medium. Each material has its unique refractive index based on its optical density.
  • For water, the refractive index is about 1.33. This means light travels slower in water compared to air, where the refractive index is 1.
  • The higher the refractive index, the more the light refracts, changing its direction more significantly.
In Snell's Law, the refractive index plays a vital role, as it directly affects the angle of refraction and incidence.
Angle of Incidence
The angle of incidence is the angle between the incident ray (the approaching light) and a line perpendicular to the boundary of two media. In the diving scenario, it's what we calculate to find how sunlight hits the water surface.
To find this, we use Snell's Law and rearrange it to solve for the angle of incidence \(\theta_1\). Snell's Law helps us by linking the angle of incidence with the angle of refraction through the refractive indices of the two mediums.
  • In our case, knowing \(\sin(\theta_1) = \frac{1}{1.33} \times \sin(28.0^\circ)\)
  • Calculating further, we find \(\theta_1\) to be approximately 20.7掳.
This calculation helps understand how light enters a medium, essential in various fields like optics and engineering.

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Most popular questions from this chapter

The near point of a naked eye is \(32 \mathrm{cm} .\) When an object is placed at the near point and viewed by the naked eye, it has an angular size of 0.060 rad. A magnifying glass has a focal length of \(16 \mathrm{cm},\) and is held next to the eye. The enlarged image that is seen is located \(64 \mathrm{cm}\) from the magnifying glass. Determine the angular size of the image.

A narrow beam of light from a laser travels through air \((n=1.00)\) and strikes point A on the surface of the water \((n=1.33)\) in a lake. The angle of incidence is \(55^{\circ} .\) The depth of the lake is \(3.0 \mathrm{m} .\) On the flat lake-bottom is point B, directly below point A. (a) If refraction did not occur, how far away from point \(B\) would the laser beam strike the lake- bottom? (b) Considering refraction, how far away from point B would the laser beam strike the lake-bottom?

An amateur astronomer decides to build a telescope from a discarded pair of eyeglasses. One of the lenses has a refractive power of 11 diopters, and the other has a refractive power of 1.3 diopters. (a) Which lens should be the objective? (b) How far apart should the lenses be separated? (c) What is the angular magnification of the telescope?

At age forty, a man requires contact lenses \((f=65.0 \mathrm{cm}\) ) to read a book held \(25.0 \mathrm{cm}\) from his eyes. At age forty-five, while wearing these contacts he must now hold a book \(29.0 \mathrm{cm}\) from his eyes. (a) By what distance has his near point changed? (b) What focal-length lenses does he require at age forty-five to read a book at \(25.0 \mathrm{cm} ?\)

A paperweight consists of a \(9.00-\mathrm{cm}-\) thick plastic cube. Within the plastic a thin sheet of paper is embedded, parallel to opposite faces of the cube. On each side of the paper is printed a different joke that can be read by looking perpendicularly straight into the cube. When read from one side (the top), the apparent depth of the paper in the plastic is \(4.00 \mathrm{cm} .\) When read from the opposite side (the bottom), the apparent depth of the paper in the plastic is \(1.63 \mathrm{cm} .\) What is the index of refraction of the plastic?
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