The figure below gives the parameter beta of the equation shownbelow versus the
ID: 1750468 • Letter: T
Question
The figure below gives the parameter beta of the equation shownbelow versus the sine of the angle theta in a two-slitinterference experiment using light of wavelength 440 nm. (The vertical axis is marked in increments of25 rad.) (a) What is the slit separation? mu m (b) Assume that none of the interference maxima are completelyeliminated by a diffraction minimum. What is the total number ofinterference maxima? (Count them on both sides of the center of theinterference pattern.) maxima (c) What is the smallest (non-negative) angle for a maximum? (d) What is the largest angle for a minimum?Explanation / Answer
to get the right answer for part d use the equation d sin()=(m+1/2) d= slit separation in part a m=((63-1)/2 )-1 so your m will equal 30 = 440nm plugging in these values gives you a =73.44 degrees the reason to subtract the extra 1 off the end of your mvalue is that if you use m=31 you will find a minimum just beyondthe diffraction minimum and not the first one between the twodiffraction minimums which is what the problem is asking for.