/*! 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 67 A point object is placed on the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 67

A point object is placed on the principal axis of a convex lens \((f=15 \mathrm{~cm})\) at a distance of \(30 \mathrm{~cm}\) from it. A glass plate \((\mu=1 \cdot 50)\) of thickness \(1 \mathrm{~cm}\) is placed on the other side of the lens perpendicular to the axis. Locate the image of the point object.

Short Answer

Expert verified
The image is formed 9.67 cm from the lens on the other side.

Step by step solution

01

Understanding the Lens Formula

The lens formula is given by \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \( f \) is the focal length, \( u \) the object distance, and \( v \) the image distance. Here, \( f = 15 \mathrm{~cm} \) and \( u = -30 \mathrm{~cm} \). The object distance is negative according to the sign convention, which considers the direction toward the lens as negative.
02

Calculate Image Distance Using Lens Formula

Substitute \( f = 15 \mathrm{~cm} \) and \( u = -30 \mathrm{~cm} \) into the lens formula: \[ \frac{1}{15} = \frac{1}{v} - \frac{1}{-30} \]Simplify: \[ \frac{1}{v} = \frac{1}{15} + \frac{1}{30} = \frac{2}{30} + \frac{1}{30} = \frac{3}{30} \]Thus, \( v = \frac{30}{3} = 10 \mathrm{~cm} \). The image is formed 10 cm from the lens.
03

Effect of Glass Plate

A glass plate of thickness \( t = 1 \mathrm{~cm} \) and refractive index \( \mu = 1.50 \) is placed on the other side. The glass plate shifts the image toward itself by a distance \( \Delta x = t\left(1 - \frac{1}{\mu}\right) \). Calculating the shift: \[ \Delta x = 1\left(1 - \frac{1}{1.50}\right) = \frac{1}{3} \mathrm{~cm} \]Thus, the apparent position of the image becomes \( 10 - \frac{1}{3} = 9.67 \mathrm{~cm} \) from the lens.
04

Conclusion

The final position of the image after considering the glass plate is 9.67 cm to the right of the lens. This accounts for the lens forming the image and the shift induced by the glass plate.

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 91影视!

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 that helps determine the relationship between the object distance, image distance, and focal length of the lens. It is written as:
  • \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
Where:
  • \( f \) is the focal length of the lens
  • \( v \) is the image distance
  • \( u \) is the object distance
When using the lens formula, it is essential to follow the sign convention in optics. For instance, distances measured towards the lens are taken as negative. Here, in our exercise:
  • \( f = 15 \) cm and \( u = -30 \) cm because the object is positioned 30 cm to the left of the lens.
The lens formula allows us to calculate where an image will be formed by the lens.
Refraction
Refraction is the bending of light as it passes from one medium to another with a different refractive index. This phenomenon occurs because light travels at different speeds in different media. An essential concept in optics, refraction is governed by Snell's Law:
  • \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \)
Where:
  • \( n_1 \) and \( n_2 \) are the refractive indices of the first and second media, respectively.
  • \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction.
In the context of the exercise, the glass plate affects the path of light after it passes through the lens. With a refractive index (\( \mu \)) of 1.50, the glass plate slightly alters the final position of the image due to refraction.
Image Distance Calculation
Calculating the image distance is crucial to finding out where the image will form using the lens formula. From the exercise:
  • The known values are \( f = 15 \text{ cm} \) and \( u = -30 \text{ cm} \).
  • Substituting these into the lens formula gives \( \frac{1}{v} = \frac{1}{15} + \frac{1}{30} \).
  • Simplifying, \( \frac{1}{v} = \frac{3}{30} \), thus \( v = \frac{30}{3} = 10 \text{ cm} \).
This means that without any other factors (like the glass plate), the image forms 10 cm from the lens. The calculation shows that the image is on the opposite side of the lens from the object, which is typical for convex lenses. The specific calculations help us arrive at this image distance.
Refractive Index
The refractive index, often denoted as \( \mu \), measures how much light bends or refracts as it enters a different medium. It is crucial in optics as it helps predict how light will behave at material boundaries. A higher refractive index indicates that light travels slower through the medium. In the given exercise:
  • The glass plate has a refractive index \( \mu = 1.50 \).
  • This value shows light travels 1.5 times slower in the glass than in a vacuum.
The refractive index of the glass affects the apparent position of the image. It shifts the image by a distance \( \Delta x = t\left(1 - \frac{1}{\mu}\right) \), where \( t \) is the thickness of the glass. In this scenario, it results in a shift of \( \frac{1}{3} \text{ cm} \), moving the image further to 9.67 cm relative to the lens position.

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

A convex lens produces a double size real image when an object is placed at a distance of \(18 \mathrm{~cm}\) from it. Where should the object be placed to produce a triple size real image ?

A hemispherical portion of the surface of a solid glass sphere \((\mu=1 \cdot 5)\) of radius \(r\) is silvered to make the inner side reflecting. An object is placed on the axis of the hemisphere at a distance \(3 r\) from the centre of the sphere. The light from the object is refracted at the unsilvered part, then reflected from the silvered part and again refracted at the unsilvered part. Locate the final image formed.

A particle goes in a circle of radius \(2 \cdot 0 \mathrm{~cm} .\) A concave mirror of focal length \(20 \mathrm{~cm}\) is placed with its principal axis passing through the centre of the circle and perpendicular to its plane. The distance between the pole of the mirror and the centre of the circle is \(30 \mathrm{~cm}\). Calculate the radius of the circle formed by the image.

A ball is kept at a height \(h\) above the surface of a heavy transparent sphere made of a material of refractive index \(\mu\). The radius of the sphere is \(R\). At \(t=0\), the ball is dropped to fall normally on the sphere. Find the speed of the image formed as a function of time for \(t<\sqrt{\frac{2 h}{g}}\). Consider only the image by a single refraction.

A \(3 \mathrm{~cm}\) tall object is placed at a distance of \(7 \cdot 5 \mathrm{~cm}\) from a convex mirror of focal length \(6 \mathrm{~cm}\). Find the location, size and nature of the image.
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Work Energy and Power

Read Explanation

Electricity

Read Explanation

Geometrical and Physical Optics

Read Explanation

Linear Momentum

Read Explanation

Scientific Method Physics

Read Explanation
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.