How Do You Calculate Slit Width Using Wavelength and Diffraction Angle?

[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?

 

[learningphysics]

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?

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)

 

[airkapp]

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 °

ahhh. thankyou

 

[learningphysics]

ahhh. thankyou

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

 

[airkapp]

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)

It is a single slit...

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

a = (3/2)λ / sin(19°)

 

[learningphysics]

It is a single slit...

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

a = (3/2)λ / sin(19°)

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