Closer. The path difference is small when the detector is far away and then increases as d decreases. We'll "get to" m = 1 first.
The intensity of a point source goes like 1/4 \pi r^2 , so if the intensity of the source is I_0, the intensity at the hole should be I_0/4 \pi x^2. Only a...
Sorry, I don't follow: m is just 1.
Yes.
Sure, but the intensity of the source is an overall constant which divides out.
I assume the source is isotropic, so no.
I guess not really, but my idea is that it's so small that it's close enough - otherwise I'm not sure what to do.
Homework Statement
A point source of light (wavelength \lambda = 600 \, \text{nm} ) is located a distance x = 10\,\text{m} away from an opaque screen with a small circular hole of radius b. A very small photodiode is moved on an axis from very far away toward the screen. The first...
One ##3##-dimensional representation of ##SO(3)## is ##3\times 3## rotation matrices parametrized by three Euler angles. The representation acts on ##3##-dimensional vectors and rotates them in ##\mathbb{R}^3 ##. That makes sense. What doesn't make sense is the interpretation of, say, the...
Problem
This is a conceptual problem from my self-study. I'm trying to learn the basics of group theory but this business of representations is a problem. I want to know how to interpret representations of a group in different dimensions.
Relevant Example
Take SO(3) for example; it's the...
Homework Statement
The picture of the circuit is attached; I want to find |V_{A}/V_{J}|. This seems really easy but I haven't done circuit analysis in forever.
Homework Equations
Complex impedances, Z_{C} = 1/i\omega C, Z_{R} = R.
The Attempt at a Solution
First R_{A} and C_{A} are in...
Ah yes, thank you! The dimensions were incorrect.
Applying the correction, we find the same result. Also, no, I didn't account for the changing magnetic field as the cylinder speeds up; doing this with Faraday yields the same answer, so it's probably right. Thanks again to both of you :smile:
Homework Statement
An infinite wire of linear charge density \lambda lies on the z axis. An insulating cylindrical shell of radius R is concentric with the wire and can rotate freely about the z axis. The charge per unit area on the cylinder is \sigma = -\lambda/2\pi R while the mass per unit...
There's a specific problem I'm doing, but this is more of a general question. The setup is a cylinder of mass m and radius R rolling without slipping down a wedge inclined at angle \alpha of mass M, where the wedge rests on a frictionless surface. I've made the Cartesian axis centred at the...
Hmm yes, that does seem to be of the same form. Honestly though, the material in the article seems way too advanced for the level that we're at (ie. the "do this because it works" stage) so I don't know if I'm even supposed to be worrying about this. You're probably right in that calling it the...
Thanks for the reply, I wasn't completely sure if this belonged in the advanced section or not. The system is an ##X_{6}## molecule with a single electron able to move between the different ##X## ions. The basis ket ##|e_{j}\rangle## represents the electron occupying the ##j^{\text{th}}## ion...
Homework Statement
If I have a Hamiltonian matrix, \mathcal{H}, that only depends on a kinetic energy operator, do the energy eigenvalues have to be non-negative? I have an \mathcal{H} like this, and some of its eigenvalues are negative, so I was wondering if they have any physical...
I have a set of data points \{\{x_1, y_1\}, \{x_2, y_2\} ... \} each with an uncertainty \{\{dx_1, dy_1,\}, \{dx_2, dy_2\} ...\}. Is there any way of fitting a nonlinear model to the data that incorporates the uncertainties on both x and y? I know that you can use the Weights command to...
Interesting. I flipped through this paper in the link and it says that "... the diﬀerential equation for the elastica, expressing curvature as a function of arclength, are equivalent to those of the motion of the pendulum..."
An unexpected fact, but disconcerting since the pendulum DE is...
Thanks, I understand what you're saying and I'm familiar with the derivation for the shape of a hanging rope, but (and I probably should have mentioned this) the wire is in a horizontal plane. Here's what it should look like at first (my sock is on the fixed end and the bent part wouldn't be...
I'm assuming you meant that moment of inertia = torque divided by angular acceleration, because what you have there isn't true (just look at the units.)
I agree with this.
I'm assuming that you're referring to this page. You'll note that T is in fact the kinetic energy of the object and...
Homework Statement
Not homework, but something I'm interested in finding out. The setup is a flexible wire with left endpoint fixed at x=0 and right endpoint at x=L. You push the right endpoint with some horizontal force directed towards the left endpoint which will move the right endpoint...
The way you've written the fractions is somewhat ambiguous. It's hard to say whether you meant that R_{\text{max}}=\frac{u^2}{g}\frac{1}{\sqrt{1+\frac{2gh}{u^2}}} or R_{\text{max}}=\frac{u^2}{g}\sqrt{1+\frac{2gh}{u^2}} (especially since the term in the radical is dimensionless), but I've...
Homework Statement
I want to estimate f(x)=\ln (\frac{1}{1+x}) on the interval (1/10,1) with the error on the approximation being no more than 0.1.
Homework Equations
http://en.wikipedia.org/wiki/Taylor's_theorem#Example
The Attempt at a Solution
Following the example from Wikipedia, I...
I'm trying to model the flight of a small spherical object (such as a ping-pong ball) through air at smallish velocities (\approx 5{_{m/s}}) with linear drag so that F_{d}=-bv. The problem is, I can't find a table of linear drag coefficients (b) anywhere; it's always just the normal drag...
You don't actually need t, that's just to confirm that there's only one root to the equation so that the collision is "just barely" avoided. The problem is still to solve this for d, which will give you the answer:
\sqrt{[-v]^2-4[(1/2)(a_{1}+a_{2})(d)]}=0
No, v can be anything. The roots are:
t = \frac{v\pm \sqrt{v^2-2d(a_{1}+a_{2})}}{a_{1}+a_{2}}
Then we picked a d to make the square root 0 so that:
t = \frac{v\pm 0}{a_{1}+a_{2}}
t = \frac{v}{a_{1}+a_{2}}
Simply eliminating one root (ie. the negative root) because a negative time doesn't make any sense is usually a good idea, but here the negativity of an answer is totally arbitrary. You could put the origin anywhere you want and even get both roots to be negative. All you know is that one car...