# Homework Help: 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.