/*! 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 77 Light is incident from glass \((... [FREE SOLUTION] | 91Ó°ÊÓ

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Light is incident from glass \((\mu=1.5)\) to water \(\left[\mu=\frac{4}{3}\right]\). Find the range of deviation for which there are two angle of incidence.

Short Answer

Expert verified
The deviation range is between 0 and 60 degrees.

Step by step solution

01

Understand Snell's Law

To determine the range of deviation where there are two angles of incidence, we need to use Snell's Law, which is given by the equation: \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \), where \( n_1 \) and \( n_2 \) are the refractive indices of the two media, and \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction, respectively.
02

Analyze the Boundary Conditions

The critical condition for having two angles of incidence is determined by the condition of total internal reflection from water back to glass. This occurs when the refracted angle \( \theta_2 \) becomes 90 degrees. At this point, we have: \( n_1 \sin(\theta_c) = n_2 \cdot 1 \). Solving for the critical angle \( \theta_c \), we find \( \theta_c = \arcsin\left(\frac{n_2}{n_1}\right) = \arcsin\left(\frac{4/3}{1.5}\right) = \arcsin\left(\frac{8}{9}\right) \).
03

Determine the Maximum Angle of Refraction

The maximum angle of refraction \( \theta_2 \) is 90 degrees. Hence, the maximum refraction condition occurs at \( \sin(\theta_2) = 1 \). Substituting this back into Snell's Law, we find that the maximum deviation condition is reached when \( \theta_1 = \theta_c \).
04

Calculate the Range of Deviation

The deviation \( \delta \) is given by \( \delta = \theta_1 - \theta_2 \). For the range where two incident angles exist, deviation \( \delta \) varies from 0 to a maximum. To ensure total internal reflection doesn't occur for any angle when the light is entering water, the deviation must be such that \( \theta_1 + \theta_2 = 180^\circ - \delta \). Therefore, \( \delta = \theta_1 + \theta_2 - 90^\circ \) leading to the range \( 0 < \delta < 90^\circ \).
05

Conclude the Range of Deviation

In this scenario, the maximum deviation that allows for two angles of incidence to occur happens when \( \delta < 60^\circ \), the remainder being constrained by the geometry of light refraction and internal reflection limits.

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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 fundamental concept in optics that defines how much a light wave bends, or refracts, when it enters a new medium. It is denoted by the symbol \( n \) and is calculated as the ratio of the speed of light in a vacuum to its speed in a given medium.
The refractive index determines how much the path of light is altered when passing between different substances. A higher refractive index means light travels slower in that medium, leading to a greater bending effect when moving between substances.
For instance, in the context of the problem, the refractive index of glass is 1.5, and for water, it is \( \frac{4}{3} \). This difference causes the light to bend or change direction as it moves from glass to water.
This concept is key in understanding Snell's Law, which defines the relationship between incident and refracted angles based on these refractive indices.
Total Internal Reflection
Total internal reflection occurs when a light ray traveling through a medium reaches the boundary of a less optically dense medium at a specific angle. Instead of refracting out into the second medium, the light is reflected entirely back into the original medium.
  • It happens only when the light travels from a medium of higher refractive index to one of lower refractive index.
  • The angle of incidence must surpass the critical angle for total internal reflection to occur, otherwise, some of the light would pass through into the new medium.
This phenomenon is significant in applications like fiber optics, where light is kept within the fibers despite bends and curves, allowing for efficient signal transmission over long distances. In the problem context, it restricts the conditions under which light can return from water to glass.
Critical Angle
The critical angle is the minimum angle of incidence within a denser medium at which light is just refracted along the boundary, rather than passing through at an angle. It’s the threshold angle for total internal reflection.
You can calculate it using the refractive indices of the two mediums:\[\theta_c = \arcsin\left(\frac{n_2}{n_1}\right)\]For example, from glass (\( n_1 = 1.5 \)) to water (\( n_2 = \frac{4}{3} \)), the critical angle is obtained from:\[\theta_c = \arcsin\left(\frac{4/3}{1.5}\right) \approx \arcsin\left(\frac{8}{9}\right)\]This calculation provides the angle at which all the light is internally reflected back into the glass, forming the basis for understanding how light manages to stay within optical devices or escape a boundary in phenomena like mirages.
Angle of Incidence
The angle of incidence is the angle at which an incoming light ray strikes a surface. Measured from the normal, an imaginary line perpendicular to the surface, it determines the behavior of the light upon interacting with the boundary.
  • A small angle of incidence might result in minimal deviation.
  • Larger angles could lead to onset conditions for phenomena like total internal reflection.
The angle of incidence is crucial in Snell’s Law:\[n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\]In solving problems involving refraction, as demonstrated, knowing your incident angle helps predict how much refraction occurs. Understanding this helps determine situations where more than one incident angle may produce similar deviations after passing through different materials.
Angle of Refraction
The angle of refraction is the angle the refracted light makes with the normal after passing into a new medium. It results from the change in speed and direction the light undergoes during the transition.
This angle is integrally connected to the refractive indices of the materials and the angle of incidence through Snell's Law:\[n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\]
Addressing the problem at hand, when light passes from glass to water, the angle of refraction is altered according to their individual refractive indices. The maximum angle of refraction is 90 degrees, which becomes critical when determining deviation conditions where total internal reflection precludes further escape and returns the light into the original medium.

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Most popular questions from this chapter

A Glass Red. Both ends of a glass rod with index of refraction \(1.60\) are ground and polished to convex hemispherical surfaces. The radius of curvature at the left end is \(6.00 \mathrm{~cm}\), and the radius of curvature at the right end is \(12.0 \mathrm{~cm}\). The length of the rod between vertices is \(40.0 \mathrm{~cm}\). The object for the surface at the left end is an arrow that lies \(23.0 \mathrm{~cm}\) to the left of the vertex of this surface. The arrow is \(1.50 \mathrm{~mm}\) tall and at right angles to the axis. (a) What constitutes the object for the surface at the right end of the rod? (b) What is the object distance for this surface? (c) Is the object for this surface real or virtual? (d) What is the position of the final image? (e) Is the final image real or virtual? Is it erect or inverted with respect to the original object? (f) What is the height of the final image?

A thin plano-convex lens fits exactly into a plane concave lens with their plane surface parallel to each other as shown in the figure. The radius of curvature of the curved surface \(R=30 \mathrm{~cm}\). The lens are made of difference material having refractive index \(\mu_{1}=\frac{3}{2}\) and \(\mu_{2}=\frac{5}{4}\) as shown in figure. (i) if plane surface of the plano-convex lens is silvered, then calculate the equivalent focal length of this system and also calculate the nature of this equivalent mirror. (ii) An object having transverse length 5 cm in placed on the axis of equivalent mirror (in part 1), at a distance \(15 \mathrm{~cm}\) from the equivalent mirror along principal axis. Find the transverse magnification produced by equivalent mirror.

A converging lens with a focal length of \(12.0 \mathrm{~cm}\) forms a virtual image \(8.00 \mathrm{~mm}\) tall, \(17.0 \mathrm{~cm}\) to the right of the lens. Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? Draw a principal-ray diagram for this situation.

An object \(0.600 \mathrm{~cm}\) tall is placed \(16.5 \mathrm{~cm}\) to the left of the vertex of a concave spherical mirror having a radius of curvature of \(22.0 \mathrm{~cm}\). (a) Draw a principal-ray diagram showing the formation of the image. (b) Determine the position, size, orientation, and nature (real or virtual) of the image.

Three thin lenses, each with a focal length of \(40.0 \mathrm{~cm}\), are aligned on a common axis; adjacent lenses are separated by \(52.0 \mathrm{~cm}\). Find the position of the image of a small object on the axis, \(80.0 \mathrm{~cm}\) to the left of the first lens.

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