/*! 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 4 The intensity produced by a poin... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The intensity produced by a point light source at a small distance \(r\) from the source is proportional to: (a) \(\frac{1}{r}\) (b) \(r^{2}\) (c) \(\frac{1}{r^{2}}\) (d) \(\frac{1}{r^{3}}\)

Short Answer

Expert verified
The intensity is proportional to \( \frac{1}{r^2} \), so option (c) is correct.

Step by step solution

01

Understanding the Concept of Light Intensity

The intensity of a point light source is the amount of light energy passing through a unit area perpendicular to the direction of light in a unit time. For a point light source, this intensity generally spreads out spherically.
02

Formulating the Proportionality

Intensity is defined as the power per unit area. As the distance from the source increases, the same amount of energy is spread over a larger spherical surface area. The surface area of a sphere with radius \( r \) is \( 4\pi r^2 \). Thus, the intensity \( I \) of light is inversely proportional to the square of the distance \( r \) from the source, written as \( I \propto \frac{1}{r^2} \).
03

Identifying the Correct Option

From the proportionality \( I \propto \frac{1}{r^2} \), the correct option for the intensity being proportional to a distance \( r \) from a point light source is option (c).

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.

Light Intensity
Light intensity is a crucial concept in physics, especially when understanding how light behaves as it spreads from a source. Imagine light shining from a flashlight in a dark room. As the beam of light moves further away, it becomes dimmer. This happens because the intensity of light diminishes with distance.
Intensity, in this context, refers to the amount of light energy that passes through a given area perpendicular to the direction of the light. The further from the source you go, the less intense the light is on a particular area. This change in intensity with distance is explained by the inverse square law. It tells us that intensity is inversely proportional to the square of the distance from the source. Thus, if you double the distance, the intensity becomes one-fourth. This can be expressed mathematically as:
- If the distance is represented by \( r \), then the intensity \( I \) is proportional to \( \frac{1}{r^2} \).Understanding this relationship helps in explaining why stars appear dimmer the further they are, despite being incredibly bright in their proximity.
Spherical Surface Area
When light radiates from a point source, it spread out in all directions forming a spherical shape around the source. This spherical distribution means that light covers a surface area that can be calculated using the formula for a sphere's surface area.
The formula is expressed as \( 4\pi r^2 \), where \( r \) is the radius of the sphere. As you increase the radius, or move further from the light source, the surface area increases dramatically. This larger surface area means that the same amount of light energy has to cover more space. Thus, each unit area of this sphere receives less energy—leading to a reduction in light intensity with increased distance.
Key points to remember about spherical surface area include:
  • The surface area increases with the square of the radius (\( r^2 \)).
  • A greater area over which the light must spread reduces its intensity on any given part of that area.
Point Light Source
A point light source is a simplified model where all the light radiates from a single, infinitesimally small point. Although actual light sources have size, understanding the concept of a point light source helps in studying light distribution and intensity without focusing on the complexities of the real-world shapes.
Point light sources help in explaining how light behaves in its most basic form. They allow us to sidestep the source's dimensions and concentrate on how light spreads and interacts with the environment. For many calculations and models in physics, the assumption of a point source is practical and sufficient.
Considering a point light source's attributes, remember:
  • It emits light uniformly in all directions.
  • It is a useful abstraction for understanding and calculating light intensity.
  • Real light sources may act like point light sources at large distances where specific size features become negligible.

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

If wavelength \(\lambda_{R}=6200 \AA\) and \(\lambda_{G}=5400 \AA\), then how fast one must rove to see red light signal as a green one? (a) \(4.1 \times 10^{7} \mathrm{~m} / \mathrm{s}\) (b) \(5.1 \times 10^{7} \mathrm{~m} / \mathrm{s}\) (c) \(6.1 \times 10^{7} \mathrm{~m} / \mathrm{s}\) (d) \(8.1 \times 10^{7} \mathrm{~m} / \mathrm{s}\)

To obtain a good photographic print, an exposure of \(2 \mathrm{~s}\) at a distance of \(1 \mathrm{~m}\) from a 75 cd bulb is done. To obtain an equally satisfactory result, what should be the distance, if time of exposure is 12 s from a 50 cd bulb? (a) \(1 \mathrm{~m}\) (b) \(2 \mathrm{~m}\) (c) \(3 \mathrm{~m}\) (d) \(4 \mathrm{~m}\)

The power of three sources \(A, B\) and \(C\) are same. The wavelengths emitted by sources are \(4300 \AA, 5550 \AA\) and \(7000 \AA\) respectively. The brightness of sources on the basis of sensation of eye are \(L_{1}, L_{2}\) and \(L_{3}\) respectively. Then: (a) \(L_{1}>L_{2}>L_{3}\) (b) \(L_{1}>L_{3}>L_{2}\) (c) \(L_{2}>L_{1}\) and \(L_{2}>L_{3}\) (d) \(L_{2}

The luminous efficiency of the lamp, if luminous flux of 200 watt lamp is 400 lumen is: (a) 2 lumen/watt. (b) 4 lumen/watt (c) 3 lumen/watt (d) 8 lumen/watt

An astronaut is approaching the moon. He sends out a radio signal of frequency \(5000 \mathrm{MHz}\) and the frequency of echo is different from that of the original frequency by \(100 \mathrm{kHz}\). His velocity of approach with respect to the moon is :(a) \(2 \mathrm{~km} / \mathrm{s}\) (b) \(3 \mathrm{~km} / \mathrm{s}\) (c) \(4 \mathrm{~km} / \mathrm{s}\) (d) \(5 \mathrm{~km} / \mathrm{s}\)

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.