/*! 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 23 An object is placed at a distanc... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An object is placed at a distance of \(3 f\) from a convex lens of focal length \(f\). A slab of refractive index \(\mu\) is placed in between lens and object. The image of the object will be formed nearest to the object if thickness of the slab is (A) \(f\) (B) \(2 f\) (C) \(\frac{f}{\mu-1}\) (D) \(\frac{\mu f}{\mu-1}\)

Short Answer

Expert verified
The thickness of the slab that would cause the image to form nearest to the object is given by (C) \[\frac{f}{\mu-1}\].

Step by step solution

01

List given information

The object distance from the lens (u) = -3f (negative sign because it is a real object) The focal length of the lens (f) = f The refractive index of the slab (μ) = μ
02

Write the lens formula

The lens formula is given as \[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\] Where, f = focal length u = object distance from the lens v = image distance from the lens (this is what we need to find)
03

Write the refractive index formula and relation with thickness

Refraction of light occurs as the light passes through the slab. The refractive index formula is given as: \[\mu = \frac{d}{t}\] Where, μ = refractive index of the slab d = distance the image appears to have moved due to refraction t = thickness of the slab Since the lens and the slab are in contact, we can establish a relationship between d and t: \[d = t(\mu - 1)\]
04

Update the object distance based on refraction

Due to the refraction caused by the slab, the image appears to form at a shifted position. Let's call the shifted object distance u': \[u' = u + d\] By substituting the formula for d from Step 3, we get: \[u' = u + t(\mu - 1)\]
05

Find the image distance nearest to the object

Using the lens formula with u' as the object distance, we can solve for the image distance v. \[\frac{1}{f} = \frac{1}{v} - \frac{1}{u'}\] \[\frac{1}{f} = \frac{1}{v} - \frac{1}{u + t(\mu - 1)}\] Rearrange the formula to solve for v: \[v = \frac{1}{\frac{1}{f} + \frac{1}{u + t(\mu - 1)}}\]
06

Minimize image distance v with respect to thickness t

Now, we want to find the thickness t that minimizes the image distance v. To do this, we take the derivative of v with respect to t and set it to zero. \[\frac{dv}{dt} = 0\]
07

Solve for the thickness t

Solve the equation from Step 6 for t: \[t = \frac{f}{\mu - 1}\] Comparing this result with the options given in the exercise, we see that the correct answer is: (C) \[\frac{f}{\mu-1}\]

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 91Ó°ÊÓ!

Key Concepts

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

Lens Formula
The lens formula is a fundamental equation in optics, providing a mathematical relationship between an object's distance from a lens, the image distance from the lens, and the lens' focal length. For a convex lens, the formula is expressed as \[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\], where the symbols represent the following:
  • \(f\) is the focal length of the lens,
  • \(v\) is the distance of the image from the lens, and
  • \(u\) is the distance of the object from the lens.
A crucial detail here is to observe the sign convention. In lens formula, distances measured in the same direction as the incoming light are positive, whereas those measured in the opposite direction are negative. Thus, real objects typically have a negative object distance (\(u\)). Improving upon the exercise further, to solve for image distance (\(v\)), you would typically rearrange the lens formula. Understanding these relationships allows one to predict where an image will be formed relative to the lens and determine properties like image size and orientation.

Refractive Index
The refractive index, often denoted as \(\(mu\)\), is a dimensionless number that describes how light propagates through a medium. It indicates the ratio of the speed of light in a vacuum to the speed of light in the medium. The refractive index formula is given by \[\(mu = \frac{c}{v}\)\], where \(c\) is the speed of light in a vacuum, and \(v\) is the speed of light in the material. In the context of the exercise involving a slab placed in front of a convex lens, it affects how the path of light is bent, or refracted, as it enters or exits the slab. The formula relating the refractive index to the thickness of the slab and the apparent shift in the object distance due to the slab is represented as \[\(d = t(\(mu\) - 1)\)\], where \(d\) is the distance by which the image appears to have moved because of refraction, and \(t\) is the actual thickness of the slab. This refraction essentially shifts the effective position of the object, which in turn impacts where the image forms through the lens. Mastering the refractive index concept is vital for solving complex optics problems, like predicting image positions when an additional medium is introduced between the object and the lens.

Optics in Physics
Optics is a branch of physics that involves the study of light, its behavior, and its properties, including its interactions with matter. It encompasses phenomena such as refraction, reflection, diffraction, and dispersion. Understanding optics is not just about knowing formulas; it's about comprehending how light travels and interacts with different mediums. In our exercise, for instance, a lens is used to focus light and form images of objects, and a slab with a particular refractive index alters the path of light rays passing through it. The observations we make in such experiments are grounded in the principles of geometric optics, where light is treated as traveling in straight lines through homogeneous media, and physical optics, which considers wave-like properties of light. By applying concepts like the lens formula and considering the refractive properties of materials, one can determine the precise optical behaviors such as the formation and position of images. This is critical in fields as varied as photography, astronomy, and vision correction. In educational contexts, simplifying these complex ideas into approachable, bite-sized lessons empowers students to build a solid foundation in the science of light.

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

The light ray is incident at angle of \(60^{\circ}\) on a prism of angle \(45^{\circ}\). When the light ray falls on the other surface at \(90^{\circ}\), the refractive index of the material of prism \(\mu\) and the angle of deviation \(\delta\) are given by (A) \(\mu=\sqrt{\frac{3}{2}}, \delta=30^{\circ}\) (B) \(\mu=1.5, \delta=15^{\circ}\) (C) \(\mu=\frac{\sqrt{3}}{2}, \delta=30^{\circ}\) (D) \(\mu=\sqrt{\frac{3}{2}}, \delta=15^{\circ}\)

We have a thin plano-convex lens of focal length \(40 \mathrm{~cm}\) and reflective index \(\mu=1.25\). It is used in various experiments as given in Column-I and results obtained are given in Column-II.(A) If we obtain virtual two times 1\. 10 magnified image, then object distance from lens (in \(\mathrm{cm}\) ) is (B) Curved surface of lens is pol- 2\. 4 ished, then lens behave like a concave mirror. Focal length of mirror (in \(\mathrm{cm}\) ) is (C) If an object is placed at a 3\. 1 distance of \(8 \mathrm{~cm}\) in front of polished lens of part (B), then distance of image from lens (in \(\mathrm{cm}\) ) is (D) If a slab of thickness \(5 \mathrm{~cm}\) and 4\. 20 refractive index \(=1.25\) is placed between object and pole perpendicular to the principal axis in part (C), then distance (in \(\mathrm{cm}\) ) by which object should be shifted to coincide image with object is

The principle of optical fibre is (A) Diffraction (B) Polarization (C) Interference (D) Total internal reflection

A thin lens of refractive index \(1.5\) has a focal length of \(15 \mathrm{~cm}\) in air. When lens is placed in a medium of refractive index (4/3), focal length will be now (A) \(30 \mathrm{~cm}\) (B) \(60 \mathrm{~cm}\) (C) \(-60 \mathrm{~cm}\) (D) \(-30 \mathrm{~cm}\)

Light of wavelength \(6328 \mathrm{~A}\) is incident normally on a slit having a width of \(0.2 \mathrm{~mm}\). The width of the central maximum measured from minimum to minimum of diffraction pattern on a screen \(9.0\) metres away will be about (A) \(0.36^{\circ}\) (B) \(0.18^{\circ}\) (C) \(0.72^{\circ}\) (D) \(0.09^{\circ}\)

See all solutions

Recommended explanations on Physics 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.