Recent content by akmphy

Please help me with this! Two deuterium nuclei overcome the Coulomb force of repulsion and attain the necessary 1 × 10−14 m distance for fusion. What is the minimum initial velocity of each nuclei, as measured in the center-of- mass frame? The permittivity of free space is 8.85419 × 10−12...

? Nuclear Fusion Homework Statement Two deuterium nuclei overcome the Coulomb force of repulsion and attain the necessary 1 × 10−14 m distance for fusion. What is the minimum initial velocity of each nuclei, as measured in the center-of- mass frame? The permittivity of free space is 8.85419 ×...

Homework Statement Light with a wavelength of 530 nm is incident on 2 slits that are spaced 1.0mm apart. If each slit has a width of .1mm, how many interference maxima lie within the central diffraction peak?Homework Equations sin( degree) = m(lambda)/d The Attempt at a Solution I have no...

Ehild, I think I understand most of what you are saying. As far as the wavelength of lambda water, are you saying that it is: lambda water= Lambda air/ n of the water That makes sense.

So far I understand that the frequency stays the same in the water, and that tmin= (lambda air)/2(n of film) I still need help on the wavelength in the water in relation to the wavelength in the air: (lambda air/ n of film)= lambda of film lambda film/2 = lambda of water = [(lambda air/n of...

Logically, of course light enters the water. But, the diagrams insinuate that the ray is reflected off the surface of the water when going from gasoline's index of refraction, to water's index of refraction. For the problem set up.. a.)how would I even begin to calculate the smallest non zero...

HELP with thin film problem! Homework Statement Consider a horizontal plane of thin film with a thickness t. This film is located in between air and water (see sketch). Light is directed from the air downward through the film and into the water, perpendicular to the surfaces.The index...

I wish I was more savvy with the computer. The angle desired is with respect to the horizontal...the water surface. However, it must be a glitch. That would be 90-31.05, which was wrong. Thanks for your help

It is an online UT program. Do you think the set up is correct? I will e-mail professor to double check.

So, I tried that... 90-31.05 = 58.95 This was wrong as well.

The problem asks for the Sun's angle of inclination? wouldn't that be the angle it strikes the surface of the water?

Angle of Inclination: ? Homework Statement An underwater scuba diver sees the Sun at an apparent angle of 23 degrees from the vertical. The refraction index for water is 1.32. what is the Sun's angle of inclination? Homework Equations n1sin(degree)= n2sin(degree) The Attempt at...

Thank you so much. I think it is right,too.

I was approaching the laser like an imaginary sphere, so I think I was using the formula of 4pir^2. So, a cross sectional area...just an area of a circle? P= 9.68e8 (.0015/2)2(pi) I get 1710.59 W. I already turned it in; is this the correct answer? Thanks for your help

So, S, intensity is 9.68e8, and the beam diameter is 0.0015m: P = AS = [(4pi)(.0015)^2][ S] = 27369.55 W I tried this one and it was wrong. I tried dividing the diameter by two to get the radius: P= AS = [4pi (7.5e-4)^2][S] = I got 6842.39 This was wrong. I tried the area of a square...