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A horizontal, parallelsided plate of glass having a refractive index of 1.52 is in contact with the surface of water in a tank. A ray coming from above in air makes an angle of incidence of 35.0\(^\circ\) with the normal to the top surface of the glass. (a) What angle does the ray refracted into the water make with the normal to the surface? (b) What is the dependence of this angle on the refractive index of the glass?

Short Answer

Expert verified
The ray makes an angle of about 35.9° with the normal in the water. It increases with the refractive index of the glass.

Step by step solution

01

Understand Snell's Law

We need to apply Snell's Law, which is the relationship between the angles of incidence and refraction, given by \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \), where \( n_1 \) and \( n_2 \) are the refractive indices of the two media, and \( \theta_1 \) and \( \theta_2 \) are the angles with respect to the normal.
02

Calculate Refraction Angle in Glass

The ray makes an angle of incidence of 35.0° in air, which has a refractive index \( n_1 = 1.0 \). Using Snell's Law, the refractive index of glass \( n_2 = 1.52 \):\[ \sin \theta_{glass} = \frac{n_1}{n_2} \sin \theta_{air} = \frac{1.0}{1.52} \sin 35.0° \]Calculate:\( \theta_{glass} = \sin^{-1} \left( \frac{1.0}{1.52} \times \sin 35.0^\circ \right) \).
03

Calculate Refraction Angle in Water

The glass-water interface refracts the ray again, now from the glass \( n_{glass} = 1.52 \) to water \( n_{water} = 1.33 \). Re-apply Snell's Law:\[ \sin \theta_{water} = \frac{n_{glass}}{n_{water}} \sin \theta_{glass} = \frac{1.52}{1.33} \sin \theta_{glass} \]Calculate:\( \theta_{water} = \sin^{-1} \left( \frac{1.52}{1.33} \times \sin \theta_{glass} \right) \).
04

Explore Dependence on Glass Refractive Index

To examine the dependence of the final angle on the refractive index of glass, consider the equation from Step 3:\[ \sin \theta_{water} = \frac{n_{glass}}{n_{water}} \sin \theta_{glass} \]Notice that \( \theta_{water} \) directly depends on both \( n_{glass} \) and \( \theta_{glass} \), which in turn depends on \( n_{glass} \) from Step 2. Hence, the angle \( \theta_{water} \) increases or decreases based on changes in the glass's refractive index \( n_{glass} \).

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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 critical concept in understanding how light behaves when it moves from one medium to another. It is denoted by the symbol \( n \) and is defined as the ratio of the speed of light in vacuum to the speed of light in the medium.

The refractive index indicates how much the path of light is bent, or refracted, when entering a material. A higher refractive index means light travels slower in that medium compared to one with a lower refractive index.

In our exercise, three different refractive indices are considered: air \( n_1 = 1.0 \), glass \( n_2 = 1.52 \), and water \( n_3 = 1.33 \). These values influence how much the light bends at each interface. By understanding and calculating refractive indices, we can predict and analyze light behaviors in various materials.
Angle of Incidence
The angle of incidence is the angle between the incoming ray of light and the normal to the surface at the point of incidence. It is a crucial factor in applying Snell's Law, which determines the path of light as it enters a new medium.

In the given problem, the angle of incidence of the light ray in the air is 35 degrees. This is the initial angle with which the light hits the first surface, which is air-glass interface.

The precise measurement of the angle of incidence is vital for predicting how much the light will bend when passing from one medium to another. These angles directly affect the subsequent refraction angles when using Snell's Law. A small change in the angle of incidence can lead to significant changes in the light's path.
Refraction Angle
The refraction angle describes the angle between the refracted ray and the normal within the medium it enters. Like the angle of incidence, it plays a fundamental role in optical calculations using Snell's Law.

In our scenario, the light passes through two interfaces: air to glass and glass to water. First, using Snell's Law on the air-glass boundary, we calculate the refracted angle in the glass. Then at the glass-water boundary, refracting again to find the angle within the water.

Each refraction angle is determined by the relationship between the refractive indices of the materials involved and the angle of incidence. Knowing these angles helps in understanding the complete path and eventual direction of the light within the final medium.
Optics
Optics is the branch of physics that deals with the study of light and its interactions with different materials. It encompasses various phenomena including reflection, refraction, diffraction, and interference.

Understanding optics involves exploring how light can be manipulated through lenses, mirrors, and other optical instruments, which all rely heavily on the principles of refraction. In the exercise scenario, optics principles are expertly applied using Snell’s Law to determine the behavior of light as it passes through multiple materials.

Thus, a strong foundation in optics not only explains natural occurrences but also equips us with the tools to design optical devices such as cameras, telescopes, and eyeglasses, by controlling light paths and focusing.
Light Behavior
The behavior of light is determined by the medium it travels through and the interfaces it encounters. When light moves from one medium to another, there are changes in speed and direction, defined by refraction and influenced by each medium's refractive index.

In the context of this exercise, light initially moves through air, then enters glass, and finally passes into water. At each boundary, it bends according to Snell’s Law. This bending is an example of refraction, which explains how we can visualize objects in different media, such as submerged objects in water that often appear bent or displaced.

The behavior of light is a key topic in physics, impacting fields from optical engineering to visual arts, demonstrating the profound influence that light's properties have on everyday experiences.

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Most popular questions from this chapter

Optical fibers are constructed with a cylindrical core surrounded by a sheath of cladding material. Common materials used are pure silica \(n_2 = 1.4502\) for the cladding and silica doped with germanium \(n_1 = 1.4652\) for the core. (a) What is the critical angle \(\theta_{crit}\) for light traveling in the core and reflecting at the interface with the cladding material? (b) The numerical aperture (NA) is defined as the angle of incidence \(theta_i\) at the flat end of the cable for which light is incident on the core-cladding interface at angle \(\theta_{crit}\) (\(\textbf{Fig. P33.46}\)). Show that sin \(\theta_i\) =\( \sqrt {n^2_1 - n^2_2}\) . (c) What is the value of \(\theta_i\) for \(n_1\) = 1.465 and \(n_2\) = 1.450?

A ray of light traveling \(in\) a block of glass (\(n\) = 1.52) is incident on the top surface at an angle of 57.2\(^\circ\) with respect to the normal in the glass. If a layer of oil is placed on the top surface of the glass, the ray is totally reflected. What is the maximum possible index of refraction of the oil?

A parallel beam of light in air makes an angle of 47.5\(^\circ\) with the surface 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 of the glass?

The vitreous humor, a transparent, gelatinous fluid that fills most of the eyeball, has an index of refraction of 1.34. Visible light ranges in wavelength from 380 nm (violet) to 750 nm (red), as measured in air. This light travels through the vitreous humor and strikes the rods and cones at the surface of the retina. What are the ranges of (a) the wavelength, (b) the frequency, and (c) the speed of the light just as it approaches the retina within the vitreous humor?

A beam of polarized light passes through a polarizing filter. When the angle between the polarizing axis of the filter and the direction of polarization of the light is \(\theta\), the intensity of the emerging beam is \(I\). If you now want the intensity to be \(I/2\), what should be the angle (in terms of \(\theta\)) between the polarizing angle of the filter and the original direction of polarization of the light?

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