/*! 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 62 A light beam of diameter \(\sqrt... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 62

A light beam of diameter \(\sqrt{3} R\) is incident symmertically on a glass hemisphere of radius \(R\) and ofrefractive index \(n=\sqrt{3}\). Find the radius of the beam at the base of hemisphere.

Short Answer

Expert verified
The beam radius at the base is \(\frac{\sqrt{3}}{2} R\).

Step by step solution

01

Understanding the Problem

We have a light beam of diameter \( \sqrt{3} R \) striking a glass hemisphere symmetrically on its curved surface. Our goal is to determine the radius of this beam as it exits at the base of the hemisphere.
02

Identify Key Concepts

Since the hemisphere has a refractive index \( n = \sqrt{3} \), Snell's law will help us understand how light bends as it enters the glass. The provided diameter implies that light enters symmetrically, focusing at the center as it passes through the hemisphere.
03

Apply Snell's Law

Snell's Law states \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \). Here, \( n_1 = 1 \) (air) and \( n_2 = \sqrt{3} \). Given the symmetrical entry and the vertical exit path, light enters at an angle that results in exiting perpendicularly.
04

Calculate Initial Entry Angle

Because light exits vertically and symmetrically at the center, initial entry angle \( \theta_1 \) is zero. With \( \sin \theta_1 = 0 \), the light does not deviate on entry, so \( \theta_2 = 0 \). Light continues unbent to the hemisphere's base.
05

Conclude Beam Exit Radius

The beam diameter remains unchanged as it enters and exits perpendicularly, so the exit diameter remains \( \sqrt{3} R \). The radius is half of this diameter.

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Key Concepts

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

light refraction
Light refraction is an interesting phenomenon that happens when light travels from one medium to another鈥攍ike from air into glass. When this occurs, the path of the light beam changes direction. This bending is due to the light's speed altering as it enters a material with a different density. A useful real-world analogy is a car moving from pavement to mud; the wheel that hits the mud first slows down, causing the car to change direction.

In the scenario of a glass hemisphere, light refraction is observed as the light beam enters the curved surface. Although the fundamental principle of Snell's Law calculates the angle of bending, in some special cases, like with symmetrical entry through the hemisphere, the light may not appear to bend visibly. Regardless of the path's appearance, refraction plays a critical role in the behavior of light as it traverses mediums.
refractive index
The refractive index is a crucial concept in determining how much a light wave bends when it transitions between materials. It is denoted as 'n' and lacks units. Simply put, it's a ratio comparing the speed of light in a vacuum to its speed in another medium. For example, the refractive index of glass might be around 1.5, indicating light travels slower in glass compared to a vacuum.
  • If a refractive index is high, it means light slows down more and thus bends more inside the medium.
  • In our problem, the refractive index of the glass is \(n = \sqrt{3}\), pointing to an interesting situation.
  • This implies that light interacts with the glass hemisphere in a specific way, shown by its symmetric entry and exit.
The refractive index helps dictate how light acts in optics, integral to designing lenses, eyeglasses, and complex optical systems.
optics
Optics is the branch of physics that deals with light and its interaction with different materials. It encompasses concepts like refraction, bending of light, and the way lenses or other optical instruments focus light. Specialized optical principles are applied in cameras, telescopes, and even eyeglasses.

Optics doesn't just cover visible light but also other electromagnetic waves like radio waves and ultraviolet light. The rules of optics are essential in understanding how light navigates surrounding environments, like moving from air into water or glass. This helps explain why objects appear bent or moved when viewed underwater or through glass.

In our exercise, optics aids in visualizing how the light beam behaves as it penetrates the glass hemisphere. By designing instruments with careful consideration of optics, scientists and engineers can enable precise applications in technology and science.

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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.

A layer of benzene \((n=1.50) 4.20 \mathrm{~cm}\) deep floats on water \((n=1.33)\) that is \(6.50 \mathrm{~cm}\) deep. What is the apparent dis. tance from the upper benzene surface to the bottom of the water laver when it is viewed at normal incidence?

A person with a near point of \(85 \mathrm{~cm}\), but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare. (a) If the lenses of the old pair have a power of \(+2.25\) diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest \(2.0 \mathrm{~cm}\) in front of his eye? (b) What would his near point be if his old glasses were contact lenses instead?

A ray of light passes through a prism in a principle plane the deviation being equal to angle of incidence which is equal to \(2 \alpha\). It is given that \(\boldsymbol{\alpha}\) is the angle of prism. Show that \(\cos ^{2} \alpha=\frac{\left(\mu^{2}-1\right)}{8}\). where \(\mu\) is the refractive index of the material of prism.
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