Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Optics and Modern Physics - Homework 3 with Solution | PHYS 275, Assignments of Optics

Material Type: Assignment; Professor: Zorba; Class: Optics & Modern Physics; Subject: Physics; University: Whittier College; Term: Unknown 1989;

Typology: Assignments

Pre 2010

Uploaded on 08/19/2009

koofers-user-iju
koofers-user-iju 🇺🇸

4.5

(2)

10 documents

1 / 3

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
PHYS 275: Hw 3 Solutions
Chapter 3:
5-) Because is the power per unit area per unit wavelength, while
is the power per unit area per unit frequency. In other words, the argument
would have been valid if the left sides of these equations were the power per unit area, but they
are not. The two formulas are not equal to each other, rather they are related by .
8-) (a)
(b)
(c) # of photons per second per second.
11-) The radiation energy per volume per unit wavelength
Since we are asked to find the number of visible photons, let us write in terms of total number
of photons per volume per wavelength by dividing the by the energy of each photon
Thus . If we integrate the right-hand side over between 400-
750 nm for room temperature, we will get , the total number of visible photons per unit volume:
.
The total number in our room would be
photons, which is almost nonexistent!
pf3

Partial preview of the text

Download Optics and Modern Physics - Homework 3 with Solution | PHYS 275 and more Assignments Optics in PDF only on Docsity!

PHYS 275: Hw 3 Solutions

Chapter 3: 5 - ) Because is the power per unit area per unit wavelength, while is the power per unit area per unit frequency. In other words, the argument would have been valid if the left sides of these equations were the power per unit area, but they are not. The two formulas are not equal to each other, rather they are related by. 8 - ) (a)  (b)  (c) # of photons per second per second. 11 - ) The radiation energy per volume per unit wavelength Since we are asked to find the number of visible photons, let us write in terms of total number of photons per volume per wavelength by dividing the by the energy of each photon Thus. If we integrate the right-hand side over between 400- 750 nm for room temperature, we will get , the total number of visible photons per unit volume: . The total number in our room would be photons, which is almost nonexistent!

Thus the fraction is 25 - ) 27 - ) 36 - ) (which is in the UV portion of the EM spectrum). 38 - ) In the Bohr model, the speed of the orbiting electron is given by. Thus the largest speed is The smallest speed is 41 - ) If gravity were the responsible force holding the atom together, we would proceed exactly the same way as in the original case except that we would replace with G , where G (6.67x , the gravitational constant, replaces the electrostatic (Coulomb) constant , and e’s would be changed by the masses of the electron and proton. Thus let us write the original expression we have derived in the class for the electrostatic case and then replace the above mentioned parameters: