Interference and Diffraction

1. May 7, 2006

prolong199

I have the follwoing question for prep for a lab, i have tried to work out the answer but i am struck with trying to work out the intensity at a certain point, in this case 4.1mm from the central maximum. I have tried using trig to calculate the angle to the point using trig then using
I=I(sin(angle/2)/(angle/2) all squared. Am i on the right track?

"..A diffraction pattern is formed on a screen 120 cm away from a 0.4 mm wide slit. Monochromatic light of 546.1 nm is used. Calculate the fractional intensity, I/I0 at a point on the screen 4.1 mm from the centre of the principal maximum..."

could some please point me in the right direction, thanks

Last edited: May 7, 2006
2. May 7, 2006

nrqed

Welcome to the forums.

This is just a matter of plugging in the numbers in an equation. The equation contains a sine squared over a certain quantity squared. Do you know this formula?

3. May 7, 2006

prolong199

sorry i forgot to add it in the first place, yes i think it is the one is added to my question?

4. May 7, 2006

nrqed

well, it depends what you mean by ''angle over 2''. What is the expression for your angle?
The angle over 2 should be equal to ${\pi a y \over L \lambda}$

5. May 7, 2006

prolong199

I worked out that the angle is 0.191 degrees, substituting into the equation i get 0.0175*I, do i then multiply by 546.1nm?, I am confused about the Io

6. May 7, 2006

nrqed

Your equation was incorrect. It should not be sin^2(angle/2)/(angle/2)^2...(this equation is the right one if by ''angle'' you mean the *phase difference* between the two waves but that is not what you are doing here).

I gave you a formula that would not require to calculate the angle, but if you want to, then use the formula

$$I = I_0 {sin^2( {\pi a sin( \theta) \over \lambda}) \over ({\pi a sin(\theta) \over \lambda})^2$$

where theta is the angle you just calculated.
A word of caution: you should put everything in radians and make sure your calculator is in radians mode when you use this formula.

Since you want the ratio I/I_0, you just need to calculate the right side of the equation and divide by I_0 so it will disappear from your result.

7. May 7, 2006

prolong199

I used trig to work out the angle from the source to the point 4.1mm from the maximum. I then used this angle in the equation.

8. May 7, 2006

nrqed

That's perfectly fine. But when y is much smaller than the distance to the wall, you can also use an equation in terms of y directly. Just replace the $sin(\theta)$ in the equation I gave before by y/L.

In any case, if you use the expression I gave in my previous post you should get the answer you are looking for.