1jus rem this
for maxima
nmax=[d/λ] [.] is gif
total maximas =2nmax+1
fo minima
nmin=[dλ+12]
total mimima =2nmin
how to find the no. of maximas in the screen if the wavelength is given?
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4 Answers
1i hv cum across ds type of ques in the fiitjee n akash papers...bt cudnt understnd the solution they gave...
13If the wavelength's given, we can find the distance b/w two consecutive maximas, let it be "S".
Let the screen width be "W".
W = n*S....then [n]+1 will give the no. of maximas.
3dsin(theta)=n(lambda) use the constraint of the sine function and find the no of maximas.
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