Easy wavelength problem

1. Mar 15, 2005

airkapp

Monochromatic light of wavelength 689 nm falls on a slit. If the angle between the first bright fringes on either side of the central maximum is 38 °, what is the slit width?

a = λ / sin(theta)

a = 689 nm / sin 38 °

a = 1.12E-3mm

is this all there is to it?

2. Mar 15, 2005

learningphysics

I believe that $$asin\theta=\lambda$$ where $\theta$ is the angle between the central line and the first bright fringe. So your theta here should be 19.

a = 689 nm / sin 19 °

EDIT: Is this a double slit or single-slit?
For double slit:
a = λ / sin(theta)

For single slit:
a = (3/2)λ / sin(theta)

Last edited: Mar 15, 2005
3. Mar 15, 2005

airkapp

ahhh. thankyou

4. Mar 15, 2005

learningphysics

Please note my edit to my previous post. Be careful whether it is a double or single slit.

5. Mar 15, 2005

airkapp

It is a single slit...

so then I do use the same theta of 19 degrees correct?

a = (3/2)λ / sin(19°)

6. Mar 15, 2005

learningphysics

Yes, that's right. theta is 19. It comes from asin(theta)=(m+1/2)λ