/*! 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 76 \(\bullet$$\bullet\) You want to... [FREE SOLUTION] | 91Ó°ÊÓ

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\(\bullet$$\bullet\) You want to support a sheet of fireproof paper horizontally, using only a vertical upward beam of light spread uniformly over the sheet. There is no other light on this paper. The sheet measures 22.0 \(\mathrm{cm}\) by 28.0 \(\mathrm{cm}\) and has a mass of 1.50 \(\mathrm{g}\) . (a) If the paper is black and hence absorbs all the light that hits it, what must be the intensity of the light beam? (b) For the light in part (a), what are the maximum values of its electric and magnetic fields? (c) If the paper is white and hence reflects all the light that hits it, what intensity of light beam is needed to support it? (d) To see if it is physically reasonable to expect to support a sheet of paper this way, calculate the intensity in a typical 0.500 \(\mathrm{mW}\) laser beam that is 1.00 \(\mathrm{mm}\) in diameter and compare this value with your answer in part (a).

Short Answer

Expert verified
a) Intensity = 4.4145×10^6 W/m^2; b) E ≈ 1834 V/m, B ≈ 6.11×10^-6 T; c) Intensity = 8.829×10^6 W/m^2; d) The laser intensity is too low to support the paper.

Step by step solution

01

Calculate Gravitational Force

The first step is to determine the gravitational force acting on the paper. The formula to calculate gravitational force is:\[ F_g = m \cdot g \]where \( m = 1.50 \text{ g} = 0.0015 \text{ kg} \) and \( g = 9.81 \text{ m/s}^2 \). Thus, \( F_g = 0.0015 \times 9.81 = 0.014715 \text{ N} \).
02

Find Required Intensity of Light to Suspend Black Paper

For part (a), the force exerted by light pressure must equal the gravitational force:\[ F_{\text{light}} = F_g \]The intensity \( I \) is given by \( I = \frac{F}{A} \), where \( A = 22.0 \text{ cm} \times 28.0 \text{ cm} = 616 \text{ cm}^2 = 0.0616 \text{ m}^2 \).Since black paper absorbs all the light, the pressure \( P = \frac{I}{c} \), where \( c \approx 3 \times 10^8 \text{ m/s} \). Hence, \[ I = F_g \cdot c = 0.014715 \times 3 \times 10^8 = 4.4145 \times 10^6 \text{ W/m}^2 \].
03

Calculate Maximum Electric and Magnetic Fields

The maximum electric field \( E \) is related to the intensity by:\[ I = \frac{1}{2} \epsilon_0 c E^2 \]where \( \epsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \). Rearranging gives:\[ E = \sqrt{\frac{2I}{\epsilon_0 c}} = \sqrt{\frac{2 \times 4.4145 \times 10^6}{8.85 \times 10^{-12} \times 3 \times 10^8}} \].Solving yields \( E \approx 1834 \text{ V/m} \).The magnetic field \( B \) is:\[ B = \frac{E}{c} \approx \frac{1834}{3 \times 10^8} \approx 6.11 \times 10^{-6} \text{ T} \].
04

Find Required Intensity for Reflective White Paper

For part (c), the force exerted by light pressure is doubled for reflective surfaces:\[ F_{\text{light}} = 2 F_g \]Intensifying the intensity:\[ I = 2 F_g \cdot c = 2 \times 4.4145 \times 10^6 = 8.829 \times 10^6 \text{ W/m}^2 \].
05

Compare with Laser Beam Intensity

For part (d), the intensity of a laser beam is calculated as:\[ I = \frac{P}{A} \]where \( P = 0.500 \text{ mW} = 0.0005 \text{ W} \) and \( d = 1.00 \text{ mm} = 0.001 \text{ m} \), so \( A = \pi \cdot (0.0005)^2 \).\[ I = \frac{0.0005}{\pi \times 0.0005^2} \approx 636.62 \text{ W/m}^2 \].Comparing, the required intensity to suspend black paper is much greater than the laser intensity.

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Key Concepts

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

Light Intensity
Light intensity refers to the amount of energy that light delivers over a specific area per unit of time. It is commonly measured in watts per square meter (W/m²). In the context of this exercise, the light intensity is vital for determining the force that light exerts on the paper. The pressure exerted by light is proportional to its intensity, thus allowing it to act against gravitational force.

To support a black sheet of paper using a light beam, the light intensity must equal the force of gravity acting on the paper divided by the area of the paper. Given the paper is black, it absorbs all the incoming light, and the required intensity is \( I = rac{F_g imes c}{A} \ \ = rac{0.014715 \, ext{N} imes 3 imes 10^8 \, ext{m/s} }{0.0616 \, ext{m}^2} \).

For a reflective surface, like white paper, the intensity required is doubled due to the law of reflection, where light returns the same amount of pressure as it initially exerted.
Gravitational Force
Gravitational force is the force with which the Earth pulls objects towards its center. It is calculated as the product of an object's mass and the acceleration due to gravity. In this exercise, the paper with a mass of 1.50 g (or 0.0015 kg) experiences a gravitational force that can be determined using the equation \( F_g = m imes g \). Here, \( g \approx 9.81 \, ext{m/s}^2 \) is the standard acceleration due to gravity on Earth.

The force calculated is fundamental in determining the light intensity required to hold up the sheet on its own without additional support. By balancing this gravitational force with the upward force exerted by light pressure, one can solve for the necessary intensity to support the paper.
Electromagnetic Fields
Electromagnetic fields are produced by moving electric charges and are part of the light's characteristics. Each light beam consists of both electric and magnetic fields, which oscillate perpendicular to each other and the direction of propagation. In this problem, calculating the maximum electric field (\( E \)) is essential for understanding the wave's energy.

The intensity \( I \) of the light is directly related to the maximum electric field by the equation \( I = \frac{1}{2} \epsilon_0 c E^2 \), where \( \epsilon_0 \) is the permittivity of free space. Knowing the intensity allows us to solve for \( E \), further deriving \( B \), the magnetic field strength, using \( B = \frac{E}{c} \).

These calculations provide insights into the electromagnetic properties of light essential for determining its practical applications, such as supporting paper.
Reflective Surfaces
Reflective surfaces, like white paper in this experiment, behave by bouncing light back instead of absorbing it, utilizing the law of reflection. This property significantly affects the force exerted by light, which is two-fold for reflective surfaces compared to absorptive ones.

For the given problem, a reflective surface doubles the effective force due to reflections, impacting the calculations for light intensity required. The intensity needed is thus \( I = 2 \, F_g \, \cdot c \), representing the paper's capability to reflect all incoming light, directing it in the opposite direction.

This underlines an important concept: the same light intensity exerted on different surfaces can yield varying results due to differences in light absorption and reflection, highlighting the intrinsic properties of material surfaces in physics.

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Most popular questions from this chapter

\(\bullet\) The intensity at a certain distance from a bright light source is 6.00 \(\mathrm{W} / \mathrm{m}^{2} .\) Find the radiation pressure (in pascals and in atmospheres) on (a) a totally absorbing surface and (b) a totally reflecting surface.

\(\bullet\) The speed of light with a wavelength of 656 \(\mathrm{nm}\) in heavy flint glass is \(1.82 \times 10^{8} \mathrm{m} / \mathrm{s} .\) What is the index of refraction of the glass at this wavelength?

\(\bullet$$\bullet\) A plane sinusoidal electromagnetic wave in air has a wave- length of 3.84 \(\mathrm{cm}\) and an \(\vec{\boldsymbol{E}}\) field amplitude of 1.35 \(\mathrm{V} / \mathrm{m}\) . (a) What is the frequency of the wave? (b) What is the \(\vec{\boldsymbol{B}}\) field amplitude? (c) What is the intensity? (d) What average force does this radiation exert perpendicular to its direction of propagation on a totally absorbing surface with area 0.240 \(\mathrm{m}^{2}\) ?

\(\bullet\) The critical angle for total internal reflection at a liquid-air interface is \(42.5^{\circ} .\) (a) If a ray of light traveling in the liquid has an angle of incidence of \(35.0^{\circ}\) at the interface, what angle does the refracted ray in the air make with the normal? (b) If a ray of light traveling in air has an angle of incidence of \(35.0^{\circ}\) at the interface, what angle does the refracted ray in the liquid make with the normal?

\(\bullet$$\bullet\) (a) Light passes through three parallel slabs of different thicknesses and refractive indexes. The light is incident in the first slab and finally refracts into the third slab. Show that the middle slab has no effect on the final direction of the light. That is, show that the direction of the light in the third slab is the same as if the light had passed directly from the first slab into the third slab. (b) Generalize this result to a stack of \(N\) slabs. What determines the final direction of the light in the last slab?

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