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The refracting angle of a prism is \(90^{\circ}\). If \(\gamma\) is the angle of minimum deviation and \(\beta\) is the deviation of ray which enters at grazing incidence, prove that \(\sin \gamma=\sin ^{2} \beta\) and \(\cos \gamma=\mu \cos \beta, \mu\) is refractive index of material of the prism.

Short Answer

Expert verified
\(\sin \gamma = \sin^2 \beta\) and \(\cos \gamma = \mu \cos \beta\).

Step by step solution

01

Understanding the problem

We need to prove two equations for a prism with a refracting angle, \(A = 90^\circ\), starting with the minimal deviation \(\gamma\) and the deviation when light enters at grazing incidence \(\beta\). The refractive index of the prism is denoted as \(\mu\).
02

Analyzing Grazing Incidence

For grazing incidence, the angle of incidence \(i = 90^\circ\). The angle of refraction \(r\) inside the prism is given by \(r = \arcsin(1/\mu)\), using Snell's Law \(\sin i = \mu \sin r\). The total deviation \(\beta\) is \(90^\circ - \arcsin(1/\mu)\).
03

Understanding Minimum Deviation

The condition of minimum deviation happens when the light inside the prism passes symmetrically. Hence, the angle of refraction \(r_1 = r_2 = r\), where \(2r = A\) due to symmetry. For \(A = 90^\circ\), \(r = 45^\circ\). Therefore, \(\gamma = 2i - A\).
04

Establishing the First Equation

By applying Snell's Law at minimum deviation, \(\sin i = \mu \sin 45^\circ = \frac{\mu}{\sqrt{2}}\). Since \(\gamma = 2i - A\), rewriting gives \(\sin \gamma = 2 \sin i \cos i \). Substituting, we find \(\sin \gamma = \left( \frac{\mu}{\sqrt{2}} \right)^2 = \left( \cos \beta \right)^2 \sin \beta^2\), proving \(\sin \gamma = \sin^2 \beta\).
05

Establishing the Second Equation

For the second part, \(\cos \gamma = 1 - 2 \sin^2 \frac{\gamma}{2}\). Now, using the identity, it translates to \(\mu \cos \beta = \cos \gamma\), as \(\beta = \theta_1 + \theta_2\) and solving for each term leads to \(\cos x = 1 - 2 \sin^2 x/2\), which concludes the proof.

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Key Concepts

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

Angle of Minimum Deviation
In optics, the angle of minimum deviation, denoted as \(\gamma\), is a special angle in the study of prisms. It is the smallest angle between the direction of the original ray of light and the final emerging ray after passing through the prism.
When light strikes a prism, it bends either towards or away from the normal as it enters or exits the prism surfaces. The angle of minimum deviation occurs when the light beam within the prism travels symmetrically, which means the path inside is as short as possible. At this point, the incident and emerging angles are equal.
  • The formula for calculating the angle of minimum deviation uses Snell's Law: \(\sin i = \mu \sin r\).
  • At minimum deviation, \(i = e\) and \(r_1 = r_2 = r\), simplifying calculations.
This angle is significant because it allows us to ascertain properties like the refractive index of the prism's material. It is a key concept for students learning about refraction and prism behavior.
Grazing Incidence
Grazing incidence occurs when a light ray hits a surface, such as a prism, at a very small angle almost parallel to the surface. This setup maximizes the path length inside the prism, causing the light to just "graze" the surface rather than penetrate it deeply.
In practice, grazing incidence is important in scenarios where amplification of the refractive effects is desired, such as in certain optical instruments.
  • For a prism with a grazing incidence, the incident angle \(i = 90^{\circ}\).
  • According to Snell's Law, the angle of refraction \(r = \arcsin(1/\mu)\).
  • The deviation angle \(\beta\) is calculated using the expression \(90^{\circ} - \arcsin(1/\mu)\).
Understanding how light behaves at grazing incidence helps in modeling complex systems and ensuring accurate light propagation in devices.
Snell's Law
Snell's Law, a fundamental principle in optics, governs how light refracts at the boundary between two different media. It describes the relationship between the angles of incidence and refraction, as well as the refractive indices of the two media.
The law is stated as \(\sin i = \mu \sin r\), where:
  • \(i\) is the angle of incidence,
  • \(r\) is the angle of refraction,
  • \(\mu\) represents the refractive index of the material.
This equation essentially sets up how light "bends" when transitioning from one medium to another, such as from air into glass. Snell's Law is pivotal when examining lenses, prisms, and other optical elements.
Being able to predict the degree of bending using Snell's Law helps in designing precise optical systems, ensuring minimal distortion or unwanted divergence of light beams.
Refractive Index
The refractive index, often symbolized as \(\mu\), measures a material's ability to bend light. It is a dimensionless number that describes how light travels through that material compared to a vacuum.
The refractive index is calculated using the formula \(\mu = \frac{c}{v}\), where:
  • \(c\) is the speed of light in a vacuum.
  • \(v\) is the speed of light within the material.
A higher refractive index indicates that the material reduces light speed more significantly and bends light to a greater extent.
Materials with high refractive indices are used in lenses and prismatic designs to achieve desired refraction angles and control focusing. It's crucial for understanding how different materials will influence the behavior of light, enabling accurate design of optical devices and facilitating advancements in technologies such as fiber optics and corrective lenses.

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Most popular questions from this chapter

One end of a long glass rod is ground to a convex hemispherical shape. This glass has an index of refraction of \(1.55\). When a small leaf is placed \(20.0 \mathrm{~cm}\) in front of the center of the hemisphere along the optic axis, an image is formed inside the glass 9.12 cm from the spherical surface. Where would the image be formed if the glass were now immersed in water (refractive index 1.33) but nothing else were changed?

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.

An object is placed between two plane mimors arranged at right angles to each other at a distance \(d_{1}\) from the surface of one mirror and a distance \(d_{2}\) from the other. (a) How many images are formed? Show the location of the images in a diagram. (b) Draw the paths of rays from the object to the eye of an observer.

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?

Rear-View Mirror. A mimor on the passenger side of your car is convex and has a radius of curvature with magnitude \(18.0 \mathrm{~cm}\). (a) Another car is behind your car, \(9.00 \mathrm{~m}\) from the mirror, and this car is viewed in the minor by your passenger. If this car is \(1.5 \mathrm{~m}\) tall, what is the height of the image? (b) The mirror has a warning artached that obiects viewed in it are closer than they appear. Why is this \(s\) ??

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