/*! 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 16 \(x-y\) plane separates two medi... [FREE SOLUTION] | 91影视

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Chapter 20: Problem 16

\(x-y\) plane separates two media. \(z>0\) contains a medium of refractive index 1 and \(z<0\) contains a medium of refractive index \(2 .\) A ray of light is incident from first medium along a vector \(\hat{i}+\hat{j}-\hat{k}\), the unit vector along refracted ray is : (a) \(\frac{1}{2 \sqrt{3}} \hat{\mathbf{i}}+\frac{1}{2 \sqrt{3}} \hat{\mathbf{j}}-\sqrt{\frac{5}{6}}^{-}\) (b) \(\frac{1}{2 \sqrt{3}} \hat{\mathbf{i}}+\frac{1}{2 \sqrt{3}} \hat{\imath}-\frac{1}{2 \sqrt{3}} \hat{k}\) (c) \(\hat{i}+\vec{j}-\hat{k}\) (d) none of the above

Short Answer

Expert verified
The correct answer is option (a).

Step by step solution

01

Identify Given Quantities

We are given two media separated by the \(x-y\) plane. The refractive index of the medium in the region \(z > 0\) is \(n_1 = 1\), and in the region \(z < 0\) the refractive index is \(n_2 = 2\). The incident ray from the first medium is along the vector \(\hat{i}+\hat{j}-\hat{k}\). Our goal is to find the unit vector direction of the refracted ray.
02

Find Direction Cosines of Incident Ray

The incident ray is along the vector \(\hat{i}+\hat{j}-\hat{k}\). Its direction cosines can be found by dividing each component by the magnitude of the vector: \[|\hat{i}+\hat{j}-\hat{k}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}\]Thus, the direction cosines are: \(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\).
03

Apply Snell's Law for Refraction

Snell's Law relates the angles of incidence \((\theta_i)\) and refraction \((\theta_r)\) with the refractive indices of the two media:\[ n_1 \sin \theta_i = n_2 \sin \theta_r \]Since the ray is given in vector form, we use the \(-k\)-component (i.e., the normal to the interface) and assume an interface normal along \(\hat{k}\). \(\sin \theta_i = \cos \theta_i = \frac{1}{\sqrt{3}}\) (since \(-\hat{k}\) component is \(-\frac{1}{\sqrt{3}}\))Substituting values:\[ 1 \cdot \frac{1}{\sqrt{3}} = 2 \cdot \sin \theta_r \]\[ \sin \theta_r = \frac{1}{2 \sqrt{3}} \]
04

Calculate Refracted Ray Direction

The refracted ray must lie in the plane formed by the incident ray and the normal to the surface. Given:- Refracted ray should have its direction reversed on the normal (\(\hat{k}\)) and maintains its coherence with the \(\hat{i}\) and \(\hat{j}\) direction.Applying direction cosines equivalent to \(\sin \theta_r \) calculated:The refracted ray direction vector is:\[ \frac{1}{2\sqrt{3}} \hat{i} + \frac{1}{2\sqrt{3}} \hat{j} - \sqrt{\frac{5}{6}} \hat{k} \]
05

Validate Solution Against Given Options

Let's compare the calculated refracted vector to the given choices: (a) \(\frac{1}{2\sqrt{3}} \hat{i} + \frac{1}{2\sqrt{3}} \hat{j} - \sqrt{\frac{5}{6}} \hat{k}\) matches our answer.Thus, option (a) is the unit vector direction of the refracted ray.

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Key Concepts

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

Snell's Law
When light travels between two different media, its speed changes, causing the light ray to bend, a phenomenon known as refraction. Snell's Law is the mathematical equation that quantifies this refraction. It states:
  • \( n_1 \sin \theta_i = n_2 \sin \theta_r \), where:
  • \( n_1 \) and \( n_2 \) are the refractive indices of the first and second media respectively.
  • \( \theta_i \) is the angle of incidence, and \( \theta_r \) is the angle of refraction.
The law shows how the angle of bending depends on the refractive indices. When light moves from a medium with a lower refractive index to a higher refractive index, the ray bends towards the normal. Conversely, it bends away from the normal when going from a higher to a lower refractive index.
Snell's Law is fundamental in understanding how lenses and other optical devices direct light rays.
Refractive Index
The refractive index, often symbolized as \( n \), is a measure of how much the speed of light is reduced inside a medium compared to the vacuum. It is calculated as:
  • \( n = \frac{c}{v} \), where:
  • \( c \) is the speed of light in a vacuum.
  • \( v \) is the speed of light in the medium.

The higher the refractive index, the slower the light travels through the material. For example, in the provided exercise, region \( z > 0 \) has a refractive index of \( 1 \), indicating light travels at its maximum speed, while region \( z < 0 \) has a refractive index of \( 2 \), causing the light to slow down significantly.
This slowing of light changes the angle at which it travels, governed by Snell's Law.
Direction Cosines
Direction cosines are utilized to describe the orientation of a vector in a 3-dimensional coordinate system. They are essentially the cosines of the angles that the vector makes with each coordinate axis.
  • To find the direction cosines of a vector, divide each component by the magnitude of the vector.
  • For a vector \( \vec{v} = a\hat{i} + b\hat{j} + c\hat{k} \), the direction cosines are \( \frac{a}{|\vec{v}|}, \frac{b}{|\vec{v}|}, \frac{c}{|\vec{v}|} \).

In the exercise, the incident ray has the vector \( \hat{i} + \hat{j} - \hat{k} \). By calculating the magnitude as \( \sqrt{3} \), the direction cosines are \( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}} \). Direction cosines ensure the proper orientation of the vector and play a critical role in the application of Snell's Law for refraction at an interface.

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Most popular questions from this chapter

Due to increase of temperature of medium, refractive index will be: (a) decreased (b) increased (c) unchanged (d) none of these

A concave mirror with its optic axis vertical and mirror facing upward is placed at the bottom of the water tank. The radius of curvature of mirror is \(40 \mathrm{~cm}\) and refractive index for water \(\mu=4 / 3\). The tank is \(20 \mathrm{~cm}\) deep and if a bird is flying over the tank at a height \(60 \mathrm{~cm}\) above the surface of water, the position of image of a bird is: (a) \(3.75 \mathrm{~cm}\) (b) \(4.23 \mathrm{~cm}\) (c) \(5.2 \mathrm{~cm}\) (d) \(3.2 \mathrm{~cm}\)

In a lake, a fish rising vertically to the surface of water uniformly at the rate of \(3 \mathrm{~m} / \mathrm{s}\), observes a bird diving vertically towards the water at a rate of \(9 \mathrm{~m} / \mathrm{s}\) vertically above it. The actual velocity of the dive of the bird is: (Given: refractive index of water \(=4 / 3\) ) (a) \(9.2 \mathrm{~m} / \mathrm{s}\) (b) \(4.5 \mathrm{~m} / \mathrm{s}\) (c) \(9.0 \mathrm{~m} / \mathrm{s}\) (d) \(3.2 \mathrm{~m} / \mathrm{s}\)

A light ray strikes a flat glass plate, at a small angle ' \(\theta^{\prime}\). The glass plate has thickness ' \(t\) ' and refractive index ' \(\mu^{\prime}\). What is the lateral displacement ' \(d^{\prime}\) ? (a) \(\frac{t \theta(\mu+1)}{\mu}\) (b) \(\frac{t \theta(\mu-1)}{\mu}\) (c) \(\frac{t}{\theta \mu}(\mu-1)\) (d) \(\frac{\mu}{t \theta}(\mu+1)\)

A lens forms a sharp image of a real object on a screen. On inserting a parallel slide between the lens and the screen with its thickness along the principal axis of the lens, it is found necessary to shift the screen parallel to itself distance \(d\) away from the lens for getting image sharply focussed on it. If the refractive index of the glass relative to air is \(\mu\), the thickness of the slab is: (a) \(\frac{d}{\mu}\) (b) \(\mu d\) (c) \(\frac{r d}{\mu-1}\) (d) \((\mu-1) \frac{d}{\mu}\)
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