Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

For four different transparent medium \(\mathrm{n}_{41} \times \mathrm{n}_{12} \times \mathrm{n}_{21}=\) (A) \(\left(1 / \mathrm{n}_{41}\right)\) (B) \(\mathrm{n}_{41}\) (C) \(\mathrm{n}_{14}\) (D) \(\left(1 / \mathrm{n}_{14}\right)\)

Short Answer

Expert verified
The correct answer is (A) \( \left(1 / \mathrm{n}_{41}\right) \).

Step by step solution

01

Understand the properties of refractive indices

Refractive index (n) is a dimensionless ratio that describes how light propagates through a medium. It is defined as the ratio of the speed of light in vacuum (c) to the speed of light in the medium (v): \( n = \frac{c}{v} \). When light travels from one medium to another, it changes its speed according to the refractive index of the second medium.
02

Determine the relationship between the given refractive indices

Let's examine the given refractive indices: n41, n12, and n21. We can rewrite these as \( n_{41} = \frac{n_4}{n_1} \), \( n_{12} = \frac{n_1}{n_2} \), and \( n_{21} = \frac{n_2}{n_1} \), where n1, n2, and n4 are the refractive indices of medium 1, medium 2, and medium 4 respectively.
03

Compute the product n41 × n12 × n21

Now, let's compute the product of the given refractive indices: \( n_{41} \times n_{12} \times n_{21} \). \[ n_{41} \times n_{12} \times n_{21} = \frac{n_4}{n_1} \times \frac{n_1}{n_2} \times \frac{n_2}{n_1} = \frac{n_4}{1} = n_{41}^{-1} \]
04

Compare the computed value with the given options

Comparing the computed value with the given options, we see that it matches option (A): \( n_{41} \times n_{12} \times n_{21} = \left(1 / \mathrm{n}_{41}\right) \) So, the correct answer is (A) \( \left(1 / \mathrm{n}_{41}\right) \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Light Propagation
Light propagation is the process by which light travels through different media. When light encounters a new medium, its speed changes due to the medium's refractive index. This change in speed causes light to bend at the interface between two media—a phenomenon known as refraction. The refractive index of a medium () is crucial for understanding this behavior.
  • The speed of light is fastest in a vacuum, with a speed denoted by c.
  • In other media, light slows down, and the ratio of c to the speed of light in the medium (v) is the refractive index: n = \( \frac{c}{v} \).
The higher the refractive index, the slower light moves in the medium. Understanding light propagation is fundamental for exploring topics like fiber optics and lens design, which rely on precise control and manipulation of light paths.
Optics
Optics is the branch of physics that studies light and its interactions with matter. It encompasses various phenomena, such as reflection, refraction, diffraction, and interference. These concepts are essential in designing optical instruments like telescopes, glasses, and microscopes. In the context of the exercise, we focus on refraction, a key aspect of optics, which occurs due to a change in light propagation speed in different materials.
Refraction leads to the bending of light rays at the boundary between different transparent media. This bending depends on the refractive indices of the involved media.
  • Snell's Law describes the relationship between angles and refractive indices when light passes through different media: \(\ n_1 \sin\theta_1 = n_2 \sin\theta_2 \).
  • Optics helps in understanding how lenses focus light and form images by utilizing refraction and reflection principles.
Thus, a deeper grasp of optics allows for better comprehension of how light can be directed and manipulated for various technological applications.
Transparent Medium
A transparent medium is a material through which light can travel with little attenuation. Such mediums include air, glass, and water, which allow light to pass while being minimally absorbed. The concept of transparency is integral to fields like optics and photonics, as it determines how well and what type of optical phenomena occur within the material.
  • In transparency, the alignment and structure of molecules often allow light waves to pass without significant scattering or absorption.
  • Refraction, reflection, and transmission are all influenced by the medium’s transparency, altering how we perceive light passing through it.
The refractive index indicates computationally how transparent a material is, with lower indices often associated with higher transparency.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Read the paragraph and chose the correct answer of the following questions In young experiment position of bright fringes is given by \(\mathrm{x}=\mathrm{n} \lambda(\mathrm{D} / \mathrm{d})\) and the position of dark fringes is given by \(\mathrm{x}=(2 \mathrm{n}-1)(N 2)(\mathrm{D} / \mathrm{d})\) where \(\mathrm{n}=1,2,3 \ldots \ldots \ldots \ldots\) for first second, third bright/dark fringe. The center of the fringe pattern is bright (for \(\mathrm{n}=0\) ). The width of each bright/dark fringe is \(\beta=(\lambda \mathrm{D} / \mathrm{d})\), Where \(\lambda=5000 \AA\). Slits are \(0.2 \mathrm{~cm}\) apart and \(\mathrm{D}=1 \mathrm{~m}\) (i) If light of wavelength \(6000 \AA\) be used in the above experiment the fringe width would be \(\mathrm{mm}\) (A) \(0.30\) (B) 3 (C) \(0.6\) (D) 6 (ii) with the light of wavelength \(5000 \AA\), If experiment were carried out under water of a \(n=(4 / 3)\) the fringe width would be (A) zero (B) \((4 / 3)\) times (C) (3/4) times (D) none of these

The width of a single slit, if the first minimum is observed at an angle of \(2^{\circ}\) with a wavelength of light \(6980 \AA\) is \(\mathrm{mm}\) (A) \(0.2\) (B) \(2 \times 10^{-5}\) (C) \(2 \times 10^{5}\) (D) \(0.02\)

A plano convex lens off \(=20 \mathrm{~cm}\) is silvered at plane surface New f will be \(\quad \mathrm{cm}\) (A) 20 (B) 40 (C) 30 (D) 60

The no. of images formed between two parallel plane mirror are (A) \(\infty\) (B) 0 (C) 180 (D) 360

Light from two coherent Sources of the same amplitude \(\mathrm{A}\) and wavelength \(\lambda\), illuminates the Screen. The intensity of the central maximum is Io. If the sources were incoherent, the intensity at the same point will be (A) \(\left(\mathrm{I}_{0} / 2\right)\) (B) \(\left(\mathrm{I}_{0} / 4\right)\) (C) \(4 \mathrm{I}_{0}\) (D) \(2 \mathrm{I}_{0}\)

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free