Diffraction Grating: 520nm & 630nm Wavelengths, 6000 Lines/cm, m=0-2 Fringes

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Homework Help Overview

The discussion revolves around a diffraction grating problem involving light wavelengths of 520nm and 630nm, with a grating of 6000 lines/cm. Participants are tasked with sketching the resulting image from m=0 to m=2, calculating angles for maxima, and determining the lowest value of m for which the 520-nm line no longer exists.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the central maximum and the relationship between wavelength and angle for different orders. There are questions about how to label the order of fringes and whether to use both wavelengths in calculations.

Discussion Status

The conversation indicates some confusion regarding the calculations and the requirements for the sketch. Participants have offered insights into the relationship between wavelength and angle, and there is ongoing exploration of how to approach the problem without a clear consensus on the methods to use.

Contextual Notes

Some participants express uncertainty about the need for a protractor in sketching and the interpretation of fringe orders. There is also mention of the condition for the 520-nm line to no longer exist, which involves the sine function exceeding 1.

rojasharma
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light wavelengths of 520nm and 630nm passes through a diffraction grating that contains 6000lines/cm. a) sketch a diagram of the image produced from m=0 to m=2. label the order of each fringe.b) calculate the angles for the first and second-order maxima that would appear on the screen. c) What is the lowest value of m for which the 520-nm line no longer exists? I have no idea how to do this question:(..
 
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i still don;t get it :(
 
rojasharma said:
i still don;t get it :(
you have all the data u want...
For (1) , there is a central maximum at theta =0 where the two rays coincide...
and the greater the wavelength , the greater the theta the ray is for the same order...
(sin function is increasing between 0 to pi/2)
(2) sintheta = nlamda/d
you have the slit separation , lamda and n , so it is possible to calculate


c) No longer exist when sin theta >1
 
the picture will be similar to the one shown in the website?..do i need to use protractor for that?
 
rojasharma said:
the picture will be similar to the one shown in the website?..do i need to use protractor for that?
sketch... I don't think you need to do that:devil:
 
label the order of each fringe? what does this mean?
 
for part b) sintheta=mlmbda/d...do i use both wavelengths? or a difference of the wavelengths?...confused:S
 

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