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Chapter 23: Problem 92

If a light beam strikes a 10 \(\mathrm{cm}\) thick slab of glass (which is immersed in air) at an angle of \(30^{\circ}\) from the normal to the surface, what will be its angle to the normal when it leaves the back side of the slab? Assume that the slab has parallel sides and an index of refraction of \(1.5 .\) $$ \begin{array}{lllllll}{\text { A. } 0^{\circ}} & {\text { B. } 30^{\circ}} & {\text { C. } 60^{\circ}} & {\text { D. } 58.3^{\circ}}\end{array} $$

Short Answer

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Step by step solution

01

Identify the given values

The angle of incidence (from the normal) is given as \(30^{\circ}\). The index of refraction of glass is \(n_1 = 1.5\) and the index of refraction of air is \(n_2 = 1.0\).
02

Apply Snell's Law for entering the glass

Snell's Law relates the angles and refractive indices of the two media: \(n_1 \sin \theta_1 = n_2 \sin \theta_2\). Plugging in the values, we compute \(1.0 \sin 30^{\circ} = 1.5 \sin \theta_2\). Solving this gives \(\sin \theta_2 = \frac{1}{3}\).
03

Calculate the refracted angle inside the glass

Use the inverse sine function to find \(\theta_2\): \(\theta_2 = \arcsin\left(\frac{1}{3}\right)\). Calculate this angle to get approximately \(19.47^{\circ}\).
04

Understand the question's geometry

Since the slab has parallel sides, the light incident on one side will exit on the other side at the same angle to the normal as it initially entered, which means the entering and exiting angles are equal when correcting for any internal refraction.
05

Conclude the angle exiting the slab

The light beam will exit at the same angle it entered the slab due to the parallel nature of the surfaces. Hence, when it leaves, the angle relative to the normal remains \(30^{\circ}\).

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Key Concepts

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

Angle of Incidence
The angle of incidence is the angle at which a light beam hits a surface. It's measured from the normal, which is an imaginary line perpendicular to the surface at the point of contact. For example, in our exercise, the light beam hits the glass slab at an angle of incidence of \(30^{\circ}\). Why is this important?
  • The angle of incidence determines how much the light beam will bend when it enters a new medium.
  • Knowing the angle of incidence helps us apply Snell's Law to find other angles in the refraction process.
Always envision the normal as a reference line that light rays deviate from as they move from one medium to another. This concept is crucial for solving many optics problems.
Refractive Index
The refractive index (or index of refraction) is a measure of how much a material slows down light passing through it compared to the speed of light in a vacuum. Every material has its own refractive index, such as air having approximately \(1.0\) and glass being \(1.5\) in this problem.
  • High refractive index indicates that light travels slower through the material, causing more bending of light rays.
  • Refractive index is essential for determining how much a light ray changes direction when entering or exiting a material.
To solve refraction problems like ours, compare the refractive indices of the two media. Through Snell's Law, you use these indices to calculate how the angles change as light travels through different substances.
Glass Slab
A glass slab is a block of glass with flat and parallel surfaces. When light enters and exits a glass slab, the angles relative to the normal at those points become key factors to consider. Because the sides of the slab are parallel:
  • Light that enters at one angle will leave at the same angle, assuming the slab does not reflect the light internally or absorb it significantly.
  • Internally, the light bends as described by Snell's Law because of the difference in refractive indices. However, upon exiting, it bends back to exit at its original incident angle.
For this exercise, the parallel sides simplify calculations, as the input and output angles are equal even if the light travels slower inside the glass.
Light Refraction
Light refraction is the phenomenon of light changing direction as it passes from one medium to another with different optical densities. This change happens due to the light's speed altering, depending on the mediating medium's refractive index.
  • When moving from a less dense to a more dense medium (like air to glass), light slows down and bends toward the normal.
  • Conversely, when exiting to a less dense medium, it speeds up and bends away from the normal.
In our example, as light enters the glass slab, it slows down and refracts. When it exits, it speeds up and refracts back, preserving the initial angle of incidence at the surface. This principle of light refraction is why lenses focus light and why objects appear distorted when viewed through water or glass.

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Most popular questions from this chapter

\(\bullet$$\bullet\) A 1.55 -m-tall fisherman stands at the edge of a lake, being watched by a suspicious trout who is 3.50 \(\mathrm{m}\) from the fisherman in the horizontal direction and 45.0 \(\mathrm{cm}\) below the surface of the water. At what angle from the vertical does the fish see the top of the fisherman's head?

\(\bullet$$\bullet\) A block of glass has a polarizing angle of \(60.0^{\circ}\) for red light and \(70.0^{\circ}\) for blue light, for light traveling in air and reflecting from the glass. (a) What are the indexes of refraction for red light and for blue light? (b) For the same angle of incidence, which color is refracted more on entering the glass?

\(\bullet\) Consider electromagnetic waves propagating in air. (a) Determine the frequency of a wave with a wavelength of (i) \(5.0 \mathrm{km},\) (ii) \(5.0 \mu \mathrm{m},\) (iii) 5.0 \(\mathrm{nm}\) . (b) What is the wavelength (in meters and nanometers) of (i) gamma rays of frequency \(6.50 \times 10^{21} \mathrm{Hz}\) , (ii) an AM station radio wave of frequency 590 \(\mathrm{kHz} ?\)

\(\bullet\) The polarizing angle for light in air incident on a glass plate is \(57.6^{\circ} .\) What is the index of refraction of the glass?

\(\bullet\) A glass plate having parallel faces and a refractive index of 1.58 lies at the bottom of a liquid of refractive index \(1.70 . \mathrm{A}\) ray of light in the liquid strikes the top of the glass at an angle of incidence of \(62.0^{\circ} .\) Compute the angle of refraction of this light in the glass.
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