/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 A flashlight beam strikes the su... [FREE SOLUTION] | 91Ó°ÊÓ

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A flashlight beam strikes the surface of a pane of glass \((n=1.56)\) at a \(63^{\circ}\) angle to the normal. What is the angle of refraction?

Short Answer

Expert verified
The angle of refraction is approximately \(34.85^{\circ}\).

Step by step solution

01

Understand Snell's Law

Snell's Law describes the relationship between the angles of incidence and refraction for a wave passing through a boundary between two different isotropic media. It is given by the formula: \[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]where \(n_1\) and \(n_2\) are the refractive indices of the first and second media, and \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction respectively.
02

Identify the Known Values

From the problem, we are given:- The angle of incidence \( \theta_1 = 63^{\circ} \).- The refractive index of air, \( n_1 = 1.00 \) (since it's not mentioned otherwise, we assume the incident medium is air).- The refractive index of glass, \( n_2 = 1.56 \).
03

Apply Snell's Law to Calculate the Angle of Refraction

Using Snell's Law, substitute the known values:\[ 1.00 \cdot \sin(63^{\circ}) = 1.56 \cdot \sin(\theta_2) \]Calculate \(\sin(63^{\circ})\) which is approximately 0.8910:\[ 0.8910 = 1.56 \cdot \sin(\theta_2) \]
04

Solve for \(\sin(\theta_2)\)

Rearrange the equation to solve for \(\sin(\theta_2)\): \[ \sin(\theta_2) = \frac{0.8910}{1.56} \]Calculate \(\sin(\theta_2)\): \[ \sin(\theta_2) \approx 0.5712 \]
05

Calculate \(\theta_2\)

Find \(\theta_2\) using the inverse sine function:\[ \theta_2 = \sin^{-1}(0.5712) \]Using a calculator, find that:\[ \theta_2 \approx 34.85^{\circ} \]
06

Confirm the Solution

Check whether the calculated values align reasonably with physical expectations: when light passes from a low refractive index medium to a higher one (air to glass), it should bend towards the normal, resulting in an angle smaller than the angle of incidence. \(34.85^{\circ}\) is less than \(63^{\circ}\), confirming that our solution is correct.

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Key Concepts

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

Angle of Incidence
In optics, the "angle of incidence" is crucial for understanding how light behaves when it hits a surface. Imagine a flashlight beam hitting a glass pane. The angle this beam makes with a perpendicular line to the surface is the angle of incidence. In the given exercise, it is specified as \(63^{\circ}\).

Thinking about angles can sometimes be confusing. An easy way to envision this is to picture a figure standing upright on the ground. The beam extends from the ground (the medium it's in, which is air in this case) upwards towards the figure's head, forming an angle at the point of contact.
  • The larger the angle, the more oblique the beam strikes the surface.
  • If you decrease the angle of incidence, the beam will approach normal incidence, hitting more directly "head-on."
Understanding this angle helps in predicting how light will behave at the boundary, such as bending towards or away from the normal depending on the refractive index if it passes through to another medium.
Angle of Refraction
When light passes from one medium to another, it doesn't just keep traveling straight—it bends. This bending process is known as refraction, and the "angle of refraction" describes the angle made by the refracted beam to the normal.

Using Snell's Law, we can calculate this angle. In the example, Snell's Law is expressed as \(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\) and helps determine the angle of refraction, \(\theta_2\).

After solving, \(\theta_2\) is found to be approximately \(34.85^{\circ}\). Notice how this is smaller than the angle of incidence. The rule of thumb to remember:
  • Light slows down and bends towards the normal when entering a denser medium.
  • Conversely, light speeds up and bends away from the normal when moving to a less dense medium.
This substratum keeps the relationship between the angle of incidence and refraction integral to understanding how light travels through various environments.
Refractive Index
The "refractive index" of a medium is a measure of how much the speed of light is reduced inside that medium. The refractive index is often denoted as \(n\) and varies between substances.

In the exercise, two refractive indices are used: for air \(n_1 = 1.00\) and for glass \(n_2 = 1.56\). These values mean light travels slower in glass compared to air. This reduction in speed causes the light to bend, changing direction at the interface.
  • The higher the refractive index, the more the light will bend when entering the material.
  • Refractive indices are determined by the material's composition and its density.
Understanding the refractive index is vital because it defines how light interacts with materials. By knowing this, we apply Snell's Law to predict the light's path as it crosses different media, thus solving problems like the one in the exercise.

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Most popular questions from this chapter

Fermat's principle states that "light travels between two points along the path that requires the least time, as compared to other nearby paths" From Fermat's principle derive \((a)\) the law of reflection \(\left(\theta_{i}=\theta_{\mathrm{r}}\right)\) and \((b)\) the law of refraction (Snell's law). [Hint: Choose two appropriate points so that a ray between them can undergo reflection or refraction. Draw a rough path for a ray between these points, and write down an expression of the time required for light to travel the arbitrary path chosen. Then take the derivative to find the minimum.]

(I) A light beam strikes a piece of glass at a \(60.00^{\circ}\) incident angle. The beam contains two wavelengths, 450.0 \(\mathrm{nm}\) and \(700.0 \mathrm{nm},\) for which the index of refraction of the glass is 1.4831 and 1.4754 , respectively. What is the angle between the two refracted beams?

(II) Some rearview mirrors produce images of cars to your rear that are smaller than they would be if the mirror were flat. Are the mirrors concave or convex? What is a mirror's radius of curvature if cars 18.0 \(\mathrm{m}\) appear 0.33 their normal size?

(II) Figure 57 shows a liquid-detecting prism device that might be used inside a washing machine or other liquid-containing appliance. If no liquid covers the prism's hypotenuse, total internal reflection of the beam from the light source produces a large signal in the light sensor. If liquid covers the hypotenuse, some light escapes from the prism into the liquid and the light sensor's signal decreases. Thus a large signal from the light sensor indicates the absence of liquid in the reservoir. If this device is designed to detect the presence of water, determine the allowable range for the prism's index of refraction \(n .\) Will the device work properly if the prism is constructed from (inexpensive) lucite? For lucite, \(n=1.5 .\)

(II) A dentist wants a small mirror that, when 2.00 \(\mathrm{cm}\) from a tooth, will produce a \(4.0 \times\) upright image. What kind of mirror must be used and what must its radius of curvature be?

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