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Chapter 23: Problem 52

Light takes \(t_{1}\) second to travel a distance \(x\) in vacuum and the same light takes \(t_{2}\) second to travel \(10 x \mathrm{~cm}\) in a medium. Critical angle for corresponding medium will be (a) \(\sin ^{-1}\left(\frac{10 t_{2}}{t_{1}}\right)\) (b) \(\sin ^{-1}\left(\frac{t_{2}}{10 t_{1}}\right)\) (c) \(\sin ^{-1}\left(\frac{10 t_{1}}{t_{2}}\right)\) (d) \(\sin ^{-1}\left(\frac{t_{1}}{10 t_{2}}\right)\)

Short Answer

Expert verified
(c) \(\sin^{-1}\left(\frac{10t_1}{t_2}\right)\) is the critical angle.

Step by step solution

01

Calculate Speed of Light in Vacuum

Given that light takes \(t_1\) seconds to travel a distance \(x\) in vacuum, the speed of light in vacuum \(c\) can be expressed as \[c = \frac{x}{t_1}.\]
02

Calculate Speed of Light in Medium

Given that the same light takes \(t_2\) seconds to travel \(10x\) cm in the medium, the speed of light in the medium \(v\) is computed as \[v = \frac{10x}{t_2}.\]
03

Determine Refractive Index of the Medium

The refractive index \(n\) of the medium is the ratio of the speed of light in vacuum to the speed in the medium: \[n = \frac{c}{v} = \frac{\frac{x}{t_1}}{\frac{10x}{t_2}} = \frac{t_2}{10t_1}.\]
04

Determine the Critical Angle

The critical angle \(\theta_c\) is given by \(\theta_c = \sin^{-1}\left(\frac{1}{n}\right)\). Substituting the expression for \(n\): \[\theta_c = \sin^{-1}\left(\frac{1}{\frac{t_2}{10t_1}}\right) = \sin^{-1}\left(\frac{10t_1}{t_2}\right).\]
05

Choose the Correct Option

From the given options, the correct expression for the critical angle is \(\sin^{-1}\left(\frac{10t_1}{t_2}\right)\), which corresponds to option (c).

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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, often denoted by the symbol \(n\), is a measure of how much a ray of light bends when it enters a medium from a vacuum. It is a crucial concept in understanding how light propagates through different materials. Given by the ratio of the speed of light in vacuum \(c\) to the speed of light in the medium \(v\), the refractive index is calculated as:
  • \( n = \frac{c}{v} \)
A higher refractive index indicates that light travels slower in the medium. Consequently, light bends more sharply upon entering the medium. This bending of light is what allows lenses to focus light, and for optical fibers to guide light over distances. Furthermore, the refractive index can be used to find critical angles, which are vital in applications like fiber optics where total internal reflection is used to trap light inside a medium.
Speed of Light in Vacuum
The speed of light in vacuum is one of the fundamental constants in physics. Represented commonly as \(c\), it is approximately \(3 \times 10^8\) meters per second. This speed is the ultimate speed limit of the universe. No information or matter can travel faster than this, according to current physical theories. In vacuum, light travels without encountering resistance, allowing it to maintain this constant velocity. Understanding this speed is essential, as it provides a baseline for comparing how light behaves in other media. When light transitions from vacuum into another medium like water or glass, it slows down, which leads to the phenomenon of refraction. Because of this crucial role, the speed of light in a vacuum is used as a reference point for calculating other optical properties such as the refractive index.
Speed of Light in Medium
When light travels through a medium other than vacuum, its speed decreases. This happens because light interacts with the particles within the medium, which causes it to slow down. The speed of light in a specific medium is denoted as \(v\) and varies depending on the medium's density and composition.For example:
  • In water, the speed of light is about \(2.25 \times 10^8\) meters per second.
  • In glass, it might drop to around \(2 \times 10^8\) meters per second.
This reduction in speed is responsible for the bending or refraction of light as it enters the surface of a new medium at an angle. Knowing this speed allows us to compute the refractive index, which tells us how much the light will bend. Hence, understanding how light slows in different media is central to the fields of optics and photonics, impacting technologies like lenses, microscopes, and even corrective eyewear.

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Most popular questions from this chapter

A plano-convex lens has a thickness of \(4 \mathrm{~cm}\). When placed on a horizontal table, with the curved surface in contact with it, the apparent depth of the bottom most point of the lens is found to be \(3 \mathrm{~cm}\). If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre of the plane face is found to be \(25 / 8 \mathrm{~cm}\). Find the foeal length of the lens. Assume thickness to be negligible (a) \(85 \mathrm{~cm}\) (b) \(59 \mathrm{~cm}\) (c) \(75 \mathrm{~cm}\) (d) \(7.5 \mathrm{~cm}\)

'The separation between the screen and a plane mirror is \(2 r\). An isotopic point source of light is placed exactly mid ways between the mirror and the screen. Assume that mirror reflects \(100 \%\) of incident light. Then the ratio of illuminance on the screen with and without the mirror is (a) \(10 ; 1\) (b) \(2: 1\) (c) \(10 ; 9\) (d) \(9: 1\)

A fish is a little away below the surface of a lake. If the critical angle is \(49^{\circ}\), then the fish could see things above water surface within an angular range of \(\theta^{\circ}\) where (a) \(\theta=49^{\circ}\) (b) \(\theta=98^{\circ}\) (c) \(\theta=24 \frac{1}{4}\) [d) \(\theta=90^{\circ}\)

A plane mirror is reflecting a ray of incident light is rotated through an angle of about an axis through the point of incidence in the plane of the mirror perpendicular to the plane of incident, then (a) the reflected ray rotates through an angle \(2 \theta\) (b) the reflected ray rotates through an angle of \(\theta\) (c) the reflected ray does not rotate (d) None of the above

1 With diaphragm the camera lens set at \(f / 2\), the correct exposure times is \(1 / 100\) s. Then with diaphragm set at \(f / 8\), the correct exposure time is (a) \(1 / 100 \mathrm{~s}\) (b) \(1 / 400 \mathrm{~s}\) (c) \(1 / 200 \mathrm{~s}\) (d) \(16 / 100 \mathrm{~s}\)
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