Wavelength of the red colour of this helium discharge tube

Click For Summary

Homework Help Overview

The discussion revolves around determining the wavelength of red light emitted from a helium discharge tube, with specific reference to the relationship between wavelength, distance, and angle in a diffraction context.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the wavelength using a formula involving distance and angle, but questions their result and seeks clarification on potential mistakes.
  • Some participants discuss the validity of the equations being used, particularly in relation to small angles and the appropriate formulas for diffraction.

Discussion Status

Participants are actively engaging with the problem, with some confirming the use of specific formulas while others are exploring different approaches. There is a collaborative atmosphere as participants offer guidance and clarification on the equations involved.

Contextual Notes

There is mention of specific values for distance and wavelength, but the original poster expresses uncertainty about their calculations and the formulas being applied, indicating a need for further exploration of the assumptions involved.

songoku
Messages
2,512
Reaction score
393
Homework Statement
A grating spectrometer is used to view the spectrum of light from a helium discharge tube. Three brightest spectral lines seen are blue, red and yellow. These lines appear at positions A, B and C in the figure but not necessarily in that order. The distance between the grating and screen is 30 cm and the slit spacing in the grating is 1870 nm. What is the wavelength of red colour?
a. 767 nm
b. 398 nm
c. 669 nm
d. 565 nm
e. 454 nm
Relevant Equations
x = λ.L / d

x = fringe separation
L = distance between grating and screen
d = slit spacing
Untitled 2.png

Red has the longest wavelength compared to the two other colours so the location of red will be at point C.

x = 11.5 cm
L = 30 cm
d = 1870 nm

Putting all into the formula, I get λ = 717 nm

Where is my msitake?

Thanks
 
Physics news on Phys.org
songoku said:
Relevant Equations:: x = λ.L / d
This equation is valid only for small angles ##\theta##.
1581384126022.png
 
Last edited:
  • Like
Likes   Reactions: songoku
is d sin θ = nλ the formula I need to use?
 
songoku said:
is d sin θ = nλ the formula I need to use?
Yes
 
  • Like
Likes   Reactions: songoku
Thank you very much for the help TSny
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
5K
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
10
Views
3K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K