Recent content by nso09

Homework Statement A player bounces a basketball on the floor, compressing it to 80.5% of its original volume. The air (assume it is essentially N2 gas) inside the ball is originally at a temperature of 20.5°C and a pressure of 1.80 atm. The ball's diameter is 23.9 cm. By how much does the...

That's true. Thank you so much for helping me out.

I see. So this is just a slightly different type of double slit problem in which the equations are switched between destructive and constructive because there is a pi shift in one of the sources, providing a relative pi phase shift.

Oh I see. So basically there is no point in the diffraction pattern where intensity is 0. The smallest intensity may be really tiny but never 0, therefore no waves will completely destructively interfere. Is that it?

Homework Statement Find at which angles θ the interference picture that appears on a distant screen made by three thin slits separated by distance d and enlightened by a source of wavelength λ (see figure) a) Shows its maxima. b) Shows its minima. multiple slit diffraction, d, ##\lambda##...

If d is less than ##\lambda##, then ##sin\theta\geq1##. So the domain doesn't fit ##sin\theta.## But how can I make use of the intensity since I don't have an angle to plug in ##I##=##I_0####((sin(\beta/2)/(\beta/2)##)^2 and set it equal to 0. Or is that the whole point? I don't have an angle...

Homework Statement Coherent electromagnetic radiation is sent through a slit of width 0.01 mm. For which of the following wavelengths will there be no points in the diffraction pattern where the intensity is zero? A. Blue light of wavelength ##\lambda=500 nm## B. Infrared light of wavelength...

Oh ok. separation between source and image is .0104m. But why do we have to take into account the image in the mirror? I'm not sure how that's relevant in the problem

Homework Statement ##d=0.52 cm,## ##\lambda=431nm##, ##L=119 m## Homework Equations ##dsin\theta=(m+1/2)\lambda## for constructive/bright fringes since there is a pi shift already The Attempt at a Solution a) $$sin\theta\leq1$$ $$(.0052/431e-9)-0.5=m=12064$$ 2 \e{3} $$ total bright fringes =...

Okay thank you. You mentioned earlier that my first attempt at a solution can't work because it made the first two waves cancel therefore there is no other wave to work with wave 3. But this time, the thin film equation is taking into account the phase difference ##2\pi/3## to make ALL THREE...

I know three waves/sources would have ##\phi=2\pi/3## so I basically have a triangle that cancels itself out. What I don't understand is when I looked at the solutions, it said that $$ 4\pi(t_1n_1)/\lambda=2\pi(m)+2\pi/3 $$ and $$ 4\pi(t_2n_2)/\lambda -\pi=2\pi(m)+2\pi/3 $$ Why is there a pi...

Homework Statement I need help on figuring out part C which is determining the ratio between ##t_1## and ##t_2## in order to get constructive interference. I'm given the index of refractions ##n_1=1.5## and ##n_2=2## and that I know that wave 1 and 2 have half cycle shift and wave 3 does not...

i would look at the x/t graph and estimate the slope of that which I noticed was negative

i know that if you have a curvy graph like y=x^2, you take the derivative of that to find the slope at that particular x. So instantaneous acceleration would be just the slope of the velocity graph if it is approximately linear?

We turned in our graphs to our TA and didn't save our data since we were so behind with our lab that we rushed. :( I'm so sorry