Diffraction, interference, single and double slit setups

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Homework Help Overview

The discussion revolves around the concepts of diffraction and interference, specifically in the context of single and double slit setups. Participants are examining calculations related to angular measurements in these setups.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are discussing the accuracy of calculations involving angular measurements and the application of approximations for small angles. There are questions about the correctness of specific values obtained and the methods used to derive them.

Discussion Status

Some participants have provided feedback on calculations, suggesting corrections and clarifications regarding the use of trigonometric functions. There is an ongoing exploration of the implications of these corrections on the results, but no consensus has been reached on the final values.

Contextual Notes

There appears to be some confusion regarding the conversion of angles and the precision of calculations, with participants referencing specific numerical values and their derivations.

Yangi
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Homework Statement
Hey,
due to the corona—virus here in Germany all schools are closed down. Today our teacher gave us two tasks but I'm not sure which formulars I should use.

1.Calculate the Angel below the 3rd.diffraction minimum behind a single slith that's 0.3 mm width. The laserpointer is a red one with a wave length of 630nm.

2.A parallel red light falls through a double slit. Interference appears on a screen that's 3meters away from the lightsource.The distance between the 1st and zero maximum is 4mm. The column spacing d=0,5mm.
Calculate the Wave length of the red light

Thanks in advance
Relevant Equations
I have added a picture with my attempts to find a solution and equations..
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I think you might have forgotten to multiply by 3 on your calculator for the first part. Also make sure to write ##\theta## in the argument of the ##\sin## function in your working! Also note that for small ##\theta##, we have ##\sin{\theta} \approx \theta## so often we can just use ##\theta = \frac{\lambda}{s}## - though in this case it's small enough that the two are the same to pretty high precision!

I agree with your part 2.
 
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Thank your for the fast reply. I corrected the first one now I got 0.30 degree is this right ?
 
Yangi said:
Thank your for the fast reply. I corrected the first one now I got 0.30 degree is this right ?

If you do want to convert it to degrees, I get a value of 0.36 degrees.
 

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