Diffraction Patterns: Calculate Wavelength of Light

[roam]

Homework Statement

A beam of monochromatic light is incident on a single slit of width 0.560 mm. A diffraction pattern forms on a wall 1.35 m beyond the slit. The distance between the positions of zero intensity on both sides of the central maximum is 2.32 mm. Calculate the wavelength of the light.

Homework Equations

y=L \ sin \theta_{dark}

sin \theta_{dark} = m \frac{\lambda}{a}

The Attempt at a Solution

sin \theta_{dark} = \frac{\lambda}{a}

And since

sin \theta_{dark} = \frac{y}{L}

We have \lambda = \frac{ya}{L} = \frac{(2.32 \times 10^{-3})(0.56 \times 10^{-3})}{1.35} = 962.37 \ nm

But why is my answer wrong?

 

[PeterO]

Homework Statement

A beam of monochromatic light is incident on a single slit of width 0.630 mm. A diffraction pattern forms on a wall 1.20 m beyond the slit. The distance between the positions of zero intensity on both sides of the central maximum is 2.28 mm. Calculate the wavelength of the light.

Homework Equations

y=L \ sin \theta_{dark}

sin \theta_{dark} = m \frac{\lambda}{a}

The Attempt at a Solution

sin \theta_{dark} = \frac{\lambda}{a}

And since

sin \theta_{dark} = \frac{y}{L}

We have \lambda = \frac{ya}{L} = \frac{(2.28 \times 10^{-3})(0.63 \times 10^{-3})}{1.2} = 1197 \ nm

But why is my answer wrong?

I haven't checked you numbers, just the idea but...

Have you taken into account the fact that formulas often work with the angle off the axis/normal to the dark fringe, where as the distance was from the dark fringe on the left to the dark fringe on the right?

 

[roam]

I haven't checked you numbers, just the idea but...

Have you taken into account the fact that formulas often work with the angle off the axis/normal to the dark fringe, where as the distance was from the dark fringe on the left to the dark fringe on the right?

Okay, I tried to do it differently, but the computer still marks me wrong:

d \ sin \theta_{min} = \lambda

tan \theta = (2.32 \times 10^{-3}){1.35} = 0.0017185

\theta = 0.09846 \ degrees

\lambda = (0.560 \times 10^{-3}) \times sin 0.09846 = 962.4 nm

What should I do?

 

[Dadface]

Look at Peters post again.

 

[Dadface]

The angle in the equation is measured from the centre of the pattern.You used a distance of 2.32mm but you should have used half of that distance.