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\(\bullet\) A beam of light in air makes an angle of \(47.5^{\circ}\) with the surface (not the normal) of a glass plate having a refractive index of 1.66 (a) What is the angle between the reflected part of the beam and the surface of the glass? (b) What is the angle between the refracted beam and the surface (not the normal) of the glass?

Short Answer

Expert verified
(a) 47.5°; (b) 66.0°.

Step by step solution

01

Understanding the Problem

We are given a light beam hitting a glass plate at an angle of \( 47.5^{\circ} \) with the surface. The refractive index of the glass is 1.66. We need to find two angles: (a) the angle between the reflected beam and the surface, and (b) the angle between the refracted beam and the surface.
02

Calculating the Incident Angle

The angle given, \(47.5^{\circ}\), is with the surface, not the normal. Hence, the incident angle (angle with the normal) \( \theta_i \) will be \( \theta_i = 90^{\circ} - 47.5^{\circ} \). Calculate \( \theta_i \): \[ \theta_i = 42.5^{\circ} \]
03

Using the Law of Reflection

According to the law of reflection, the angle of incidence is equal to the angle of reflection. Therefore, the angle of reflection \( \theta_r \) with the normal is:\[ \theta_r = \theta_i = 42.5^{\circ} \] To find the angle with the surface, use \( 90^{\circ} - \theta_r \):\[ \angle_{reflection} = 90^{\circ} - 42.5^{\circ} = 47.5^{\circ} \]
04

Applying Snell's Law for Refraction

Snell's Law is given by:\[ n_1 \sin(\theta_i) = n_2 \sin(\theta_t) \]where \( n_1 = 1 \) (air), \( n_2 = 1.66 \) (glass), and \( \theta_t \) is the angle of refraction.Substitute the known values:\[ 1 \cdot \sin(42.5^{\circ}) = 1.66 \cdot \sin(\theta_t) \]Solve for \( \sin(\theta_t) \):\[ \sin(\theta_t) = \frac{\sin(42.5^{\circ})}{1.66} \]
05

Calculating the Angle of Refraction

Calculate \( \sin(42.5^{\circ}) \) which is approximately 0.674:\[ \sin(\theta_t) = \frac{0.674}{1.66} \approx 0.406 \]Now calculate \( \theta_t \) by finding the inverse sine:\[ \theta_t \approx \sin^{-1}(0.406) \approx 24.0^{\circ} \]
06

Finding the Angle with the Surface

The angle between the refracted beam and the surface is:\[ \angle_{refraction} = 90^{\circ} - \theta_t = 90^{\circ} - 24.0^{\circ} = 66.0^{\circ} \]

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Key Concepts

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

Law of Reflection
The Law of Reflection is a fundamental concept in physics that describes how light behaves when it encounters a surface. It states that the angle of incidence, which is the angle the incoming ray makes with the normal (an imaginary line perpendicular to the surface), is equal to the angle of reflection. This principle applies to smooth surfaces, causing reflections like those seen in mirrors.
For example, in our problem, the light strikes the glass plate at an angle of incidence of 42.5° with the normal. According to the Law of Reflection, the angle of reflection will also be 42.5°. This means the reflected light behaves predictably when it bounces off surfaces. This principle is crucial for understanding how light behaves in optics.
Refractive Index
The refractive index is a measure of how much a medium bends or refracts light. It's a dimensionless number that describes how fast light travels in a medium compared to a vacuum. For instance, in air, the refractive index is approximately 1, while in denser materials like glass, it is higher.
In the problem, the glass has a refractive index of 1.66, indicating that light travels slower in glass than in air. This difference in speed causes the light to change direction when it enters or leaves the glass. Understanding refractive index is crucial for fields like optics, where precision in light manipulation is key.
Angle of Incidence
The angle of incidence is the angle between the incoming light ray and the normal to the surface. It's integral in both reflecting and refracting light as it dictates the initial path of the light.
In our exercise, the angle given was with the surface, so it was necessary to convert this to an angle with the normal by subtracting from 90°. The original angle with the surface was 47.5°, leading to an angle of incidence of 42.5°. This angle is used to determine the angle of reflection owing to the Law of Reflection. Additionally, Snell's Law employs the angle of incidence to calculate the angle of refraction.
Angle of Refraction
The angle of refraction is the angle between the refracted ray and the normal inside the medium. It varies depending on the refractive indices of the two media involved, as described by Snell's Law.
Snell's Law is formulated as: \[ n_1 \sin(\theta_i) = n_2 \sin(\theta_t) \]where \( n_1 \) and \( n_2 \) are the refractive indices of the initial and second medium, and \( \theta_i \) and \( \theta_t \) are the angles of incidence and refraction respectively. In our case, solving Snell's Law gave an angle of refraction of approximately 24.0° with the normal. This showcases how light enters a denser medium, slowing down and bending towards the normal. Understanding this angle is key to comprehending light behavior as it moves between different substances.

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Most popular questions from this chapter

If the entire slab (without the oil) is submerged in a fluid with an index of refraction of \(1.5,\) what will be the effect? A. The slab will appear to change color. B. Light striking the slab could be totally reflected. C. The slab will be very difficult to see. D. Light exiting the slab could be totally reflected.

\(\bullet\) Unpolarized light with intensity \(I_{0}\) is incident on an ideal polarizing filter. The emerging light strikes a second ideal polarizing filter whose axis is at \(41.0^{\circ}\) to that of the first. Determine (a) the intensity of the beam after it has passed through the second polarizer and (b) its state of polarization.

\(\bullet\) (a) How much time does it take light to travel from the moon to the earth, a distance of \(384,000 \mathrm{km} ?\) (b) Light from the star Sirius takes 8.61 years to reach the earth. What is the distance to Sirius in kilometers?

\(\bullet$$\bullet\) An optical fiber consists of an outer "cladding" layer and an inner core with a slightly higher index of refraction. Light rays entering the core are trapped inside by total internal reflection and forced to travel along the fiber (see Figure \(23.59 ) .\) Suppose the cladding has an index of refraction of 1.46 and the core has an index of refraction of \(1.48 .\) Calculate the largest angle \(\theta\) between a light ray and the longitudinal axis of the fiber ( see the figure) for which the ray will be totally internally reflected at the core/cladding boundary.

\(\bullet$$\bullet\) A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area 0.500 \(\mathrm{m}^{2} .\) At the window, the electric field of the wave has rea value 0.0200 \(\mathrm{V} / \mathrm{m} .\) How much energy does this wave carry through the window during a 30.0 s commercial?

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