Diffraction Patterns: Calculate Wavelength of Light

AI Thread Summary
A beam of monochromatic light creates a diffraction pattern through a single slit, with specific measurements provided for slit width and distance to the wall. The initial calculations for the wavelength of light resulted in values around 962.37 nm and 1197 nm, but both were marked incorrect. The discussion highlights the importance of using the correct distance for calculations, specifically noting that the angle in the equations should be measured from the center of the pattern, requiring half the distance between dark fringes. Adjusting the calculations accordingly leads to the correct wavelength determination. Accurate application of diffraction formulas is essential for obtaining valid results in such problems.
roam
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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? :confused:
 
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roam said:

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? :confused:

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?
 
PeterO said:
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?
 
Look at Peters post again.
 
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.
 
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