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Relation between critical angle of water \(C_{w}\) and that of the glass \(\mathrm{C}_{\mathrm{g}}\) is (given, \(\left.\mathrm{n}_{\mathrm{w}}=(4 / 3), \mathrm{n}_{\mathrm{g}}=1.5\right)\) (A) \(\mathrm{C}_{\mathrm{w}}<\mathrm{C}_{\mathrm{g}}\) (B) \(\mathrm{C}_{\mathrm{w}}=\mathrm{C}_{\mathrm{g}}\) (C) \(\mathrm{C}_{\mathrm{w}}>\mathrm{C}_{\mathrm{g}}\) (D) \(\mathrm{C}_{\mathrm{w}}=\mathrm{C}_{\mathrm{g}}=\mathrm{O}\)

Short Answer

Expert verified
The relationship between the critical angles of water and glass is \(C_\textrm{w} > C_\textrm{g}\), as the critical angle for water (\(\arcsin\left(\frac{3}{4}\right)\)) is greater than the critical angle for glass (\(\arcsin\left(\frac{2}{3}\right)\)). Thus, the correct option is (C).

Step by step solution

01

Write down the formula for the critical angle

The critical angle formula based on Snell's law is: \( C = \arcsin\left(\frac{n_2}{n_1}\right) \) where,\(C\) is the critical angle; \(n_1\) is the refractive index of the first medium; \(n_2\) is the refractive index of the second medium.
02

Calculate the critical angles for water and glass

We are given the refractive indices for water and glass respectively: \(n_\textrm{w}= \frac{4}{3}\) (water) \(n_\textrm{g}= 1.5\) (glass) As light moves from water and glass to the air (refractive index of air is approximately 1), we can calculate the critical angles for water and glass by plugging in the refractive indices into the critical angle formula. For water: \(C_\textrm{w} = \arcsin\left(\frac{n_\textrm{air}}{n_\textrm{w}}\right) = \arcsin\left(\frac{1}{\frac{4}{3}}\right) \) For glass: \(C_\textrm{g} = \arcsin\left(\frac{n_\textrm{air}}{n_\textrm{g}}\right) = \arcsin\left(\frac{1}{1.5}\right) \)
03

Find the relationship between the critical angles of water and glass

Calculate the critical angles and compare them to find the relationship: \(C_\textrm{w} = \arcsin\left(\frac{3}{4}\right)\) \(C_\textrm{g} = \arcsin\left(\frac{2}{3}\right)\) Now compare the values: \( C_\textrm{w} > C_\textrm{g} \) The comparison shows that the critical angle of water is greater than the critical angle of glass.
04

Choose the correct option

Based on the solution, the correct option is: (C) \(C_\textrm{w} > C_\textrm{g}\)

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Key Concepts

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

Snell's Law
Snell's Law is fundamental when studying the behavior of light as it moves between different media. It relates the angles of incidence and refraction to the refractive indices of the two substances involved. The core formula is:\[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \]where:
  • \(n_1\) and \(n_2\) are the refractive indices of the first and second medium, respectively.
  • \(\theta_1\) is the angle of incidence, and \(\theta_2\) is the angle of refraction.
This law helps predict how light will bend when transitioning from one medium to another, such as air to water or glass. It is particularly useful in understanding the concept of the critical angle, which occurs when light passes from a medium with a higher refractive index to one with a lower refractive index, like water to air. At this point, light refracts at an angle of 90 degrees along the boundary itself. Understanding Snell's Law allows us to appreciate the optical phenomena of refraction and total internal reflection, which are key in designing optical devices like lenses and prisms.
Refractive Index
The refractive index is a measure of how much light slows down as it enters a material. It is a dimensionless number and is crucial in determining how much light will bend, or refract, when entering a new medium. The refractive index \(n\) of a material is defined by:\[ n = \frac{c}{v} \]where:
  • \(c\) is the speed of light in a vacuum (approximately \(3 \times 10^8\) meters per second).
  • \(v\) is the speed of light in the material.
A higher refractive index indicates that light travels more slowly in the medium, which usually results in greater bending of light upon entering the material. For instance, glass and water both have higher refractive indices than air, causing a noticeable bending of light. This concept is essential for calculating the critical angle through Snell's Law, allowing us to determine the conditions under which total internal reflection occurs.
Optics
Optics is a branch of physics that studies the behavior and properties of light, including its interactions with matter. It encompasses various phenomena such as reflection, refraction, diffraction, and interference. Understanding optics is critical for developing technologies like lenses, mirrors, telescopes, and fiber optics. Refraction is a pivotal part of optics, illustrating how light bends at the interface between two different media. When light moves from a medium like water or glass into air, the change in speed and direction is predictable using the refractive indices of the materials involved. One interesting aspect of optics is the critical angle and total internal reflection. These principles are not only theoretical but have practical applications in designing optical fibers for telecommunications. By keeping light confined via total internal reflection, fibers can transmit data over long distances efficiently. Learning optics helps us understand both natural and engineered phenomena, from the beauty of rainbows to the precision of laser technology. As light moves through different media, its behavior continues to intrigue and inspire advances in technology and science.

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Most popular questions from this chapter

$$ \begin{array}{|l|l|} \hline \text { Column - I } & \text { Column - II } \\ \hline \text { (i) Minimum deviation } & \text { (a) }(\mathrm{n}-1) \mathrm{A}+\left(\mathrm{n}^{\prime}-1\right) \mathrm{A}^{\prime}=0 \\ \text { (ii) Angular dispersion } & \text { (b) }\left[\left(\mathrm{n}_{\mathrm{v}}-\mathrm{n}_{\mathrm{r}}\right) /(\mathrm{n}-1)\right] \\ \text { (iii) Dispersive power } & \text { (c) }\left(\mathrm{n}_{\mathrm{v}}-\mathrm{n}_{\mathrm{r}}\right) \mathrm{A} \\ \text { (iv) Condition for no deviation } & \text { (d) }(\mathrm{n}-1) \mathrm{A} \\ \hline \end{array} $$ (A) \(i-c\), ii \(-d\), ii \(-b\), iv-a (B) \(i-a, 1 i-b\), iii \(-c\), iv-d (C) \(i-c\), ii \(-b\), ii \(-a\), iv-d (D) \(\mathrm{i}-\mathrm{d}, \mathrm{i} i-\mathrm{c}, \mathrm{ii}-\mathrm{b}, \mathrm{iv}-\mathrm{a}\)

A concave lens forms the image of an object such that the distance between the object and the image is \(10 \mathrm{~cm}\) and the magnification produced is \((1 / 4)\), the focal length of lens will be \(\mathrm{cm}\) (A) - 6.2 (B) \(-12.4\) (C) \(-4.4\) (D) \(-8.8\)

A ray of light passes through a prism having refractive index \((\mathrm{n}=\sqrt{2})\), Suffers minimum deviation If angle of incident is double the angle of refraction within prism then angle of prism is (A) \(30^{\circ}\) (B) \(60^{\circ}\) (C) \(90^{\circ}\) (D) \(180^{\circ}\)

A ray of light from denser medium strikes a rarer medium at angle of incidence i. The reflected and refracted rays make an angle of \(90^{\circ}\) with each other. The angle of reflection and refraction are r and \(\mathrm{r}^{\prime}\) respectively. The critical angle is (A) \(\sin ^{-1}(\tan i)\) (B) \(\tan ^{-1}(\tan \mathrm{r})\) (C) \(\tan ^{-1}(\sin i)\) (D) \(\sin ^{-1}(\tan \mathrm{r})\)

A mark at the bottom of the liquid appears to rise by \(0.2 \mathrm{~m}\), If depth of the liquid is \(2.0 \mathrm{~m}\) then refractive index of the liquid is (A) \(1.80\) (B) \(1.60\) (C) \(1.33\) (D) \(1.11\)

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