# Young'a double slit experiment

1. Mar 26, 2007

### kokok

1. The problem statement, all variables and given/known data
in a double-slit experiment, blue light of wavelength 4.60*10^2 nm gives a second order maximum or CI at a certain location P on the screen. what wavelength of visible light would have a minimum or DI at P?

2. Relevant equations
x=(l*lamda)/separation

3. The attempt at a solution
I dun get it at all

2. Mar 26, 2007

### 21385

The equation for double slit experiments when it is a maximum is
sin A=m(wavelength)/d. The equation for minimums is
sin A=(m+0.5)(wavelength)/d. Where A is the angle from where the observer will look at the spot on the screen. M is the "order" of the spots, whether it be max or min. d is the distance between the slits. Put all the constants on one side, which will be the sin A and d. Simple math and equating the equations together will get you the correct answer.

The question is a bit of vague, since it does not say what order is the minimum.

3. Mar 26, 2007

### kokok

i dunno it was nelson's text book chapter 9 review Qs

4. Mar 26, 2007

### Staff: Mentor

But it does say visible light.

5. Mar 26, 2007

### 21385

lol, i missed that ;)

6. Mar 26, 2007

### kokok

um..how should i start it...

7. Mar 26, 2007

### kokok

How should i start it!?><

8. Mar 26, 2007

### 21385

well, just follow the instructions in my previous post

move the constants to one side for both equations and equate the other sides together.

m1(wavelength1)=(m2+0.5)(wavelength2)
you know that m1 is 2 and wavelength1 is 4.60*10^2 nm.
Now try different values for m2 that will result in a wavelength that's between 400 to 700 nm. (visible light)

9. Mar 26, 2007

### kokok

oh, so when i replaced m2 to 1, i got about 600nm..i think thatz the answer.