Optics: I do not understand what this question is asking

[Xyius]

Homework Statement

Call the irradiance at the center of the central Fraunhofer diffraction maximum of a single slit $I_0$ and the irradiance at some other point in the patterm I. Obtain the ratio $I/I_0$ for some point on the screen that is 3/4 of a wavelength farther from one edge of the slit than the other.

Homework Equations

Irradiance for single slit diffraction
$$I=I_0sinc^2(β)$$
$$β=\frac{1}{2}kbsin\theta$$
$$k=\frac{2\pi}{\lambda}$$
b=slit size

The Attempt at a Solution

Well I know that when beta is equal to zero, the irradiance is at its maximum. But I do not know if this helps me or not. I honestly do not understand the question. 3/4 of a wavelength farther from one edge of the slit than the other? Don't I need the slit length? How can I make use of this information? Why does it say "than the other." What other point?

I really want to understand this problem. If anyone can help me I would much appreciate it!

 

[Andrew Mason]

The Attempt at a Solution

Well I know that when beta is equal to zero, the irradiance is at its maximum. But I do not know if this helps me or not. I honestly do not understand the question. 3/4 of a wavelength farther from one edge of the slit than the other? Don't I need the slit length? How can I make use of this information? Why does it say "than the other." What other point?

I really want to understand this problem. If anyone can help me I would much appreciate it!

Think of two coherent light sources, one at each edge of the slit. At a point that is 3/4 of a wavelength farther from one of those two sources than the other, how would the amplitude of the combined wave compare to the amplitude at a point where there was full constructive interference (maximum amplitude). How would the intensities be related? (ie how is intensity related to amplitude?).

AM

 

[netgypsy]

In other words you don't need a variable if you have a way to cancel it using the given information and perhaps setting up two equations or a ratio

 

[Xyius]

Ohh! Okay that makes sense now! Thanks guys, I will give it a shot tomorrow. (Just finished studying :p)