Solving Destructive Interference: Angle vs Wavelength

Click For Summary
The discussion focuses on the destructive interference equation for small angles, specifically angle = wavelength/(2a), where 'a' is the width. A user initially misapplied a double slit equation instead of the correct single slit equation, leading to a significant error in their wavelength calculation. After reviewing the relevant physics resource, they recognized the need to account for an additional factor of 2 and clarified their variables. By correctly applying the single slit equation, they recalculated the width and achieved a more accurate understanding of the interference patterns. The conversation emphasizes the importance of using the correct equations in physics problems.
JoeyBob
Messages
256
Reaction score
29
Homework Statement
See attached
Relevant Equations
angle=wavelength/(2a)
The destructive interference equation for small angles is angle=wavelength/(2a), where a is the width. I assume it means destructive interference since its talking about areas where no light is present.

Using the equation after changing degrees into radians I get the answer of 2491 nm when the answer shoould be 9982 nm. the answer is approx. 4 times as large. Where am I going wrong here?
 

Attachments

  • question.PNG
    question.PNG
    7.3 KB · Views: 145
Physics news on Phys.org
BvU said:
Hi,

Study http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html to find a factor of 2 (basically: you have the wrong relevant equation) and consider that the exercise text (were you not to be bothered retyping it ?) talks about the second angle to find another factor of 2 ...

##\ ##
i see, I was using a double slit eqn instead of a single slit. Looking at your link, tanx=x=y/D.

So i don't know what y is nor d.

d=y/x=y/0.108 = 9.2593y

Now I can use y=(m*wavelength*D)/a to find width. m i assume is 2 because second angle measurement...

0.108d=(2*539 nm *d)/a

a=9981.4815

This is good. Thanks. I have more understanding over slits now that *hopefully* I won't just haphazardly use an equation that seems appropriate.
 
  • Like
Likes BvU and berkeman
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 3 ·
Replies
3
Views
898
  • · Replies 13 ·
Replies
13
Views
5K
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
1
Views
3K
Replies
7
Views
6K
  • · Replies 6 ·
Replies
6
Views
922