Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Introductory Physics Homework Help
Interference Fringes in a Diffraction envelope
Reply to thread
Message
[QUOTE="Charles Link, post: 6058156, member: 583509"] Usually the letter ## b ## us used for slit width. The diffraction envelope has its zeros when ## m \lambda=b \sin(\theta) ## with integer ## m \neq 0 ##. The question is asking how many maxima do you find between the ## \theta ## corresponding to ## m=-1 ## and ## m=+1 ##. The maxima of the interference pattern are found where ## m \lambda=d \sin(\theta) ##, and this time integer ## m=0 ## is included. Here ## d ## is the distance between slits. ## \\ ## You need to find the largest ## m=m_c ## where ## \theta ## is still less than the place where the diffraction envelope first goes to zero that you found above. Then to count how many maxima you have including ## m=0 ## inside that envelope. That count is a very simple ## 2m_c+1 ##. ## \\ ## And let me give you what may be a helpful hint: I think you will find that ## m=8 ## of the second part lands right on the ## m=1 ## of the first part. Here, that maximum will not appear because the diffraction envelope gets multiplied by the interference pattern intensity result (for very narrow slits or point sources), giving zero intensity there. It should be fairly obvious what ## m_c ## is. And note that the wider the slit, the narrower the diffraction envelope becomes. Point sources and thin slits have very wide diffraction envelopes. [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Introductory Physics Homework Help
Interference Fringes in a Diffraction envelope
Back
Top