1is dis the fulll qn?
35 Answers
24give the soln.....................
i am struggling with this for so many daysssss
1in this q just divide the circle in 4 equal parts like cordinate axis n origin kkkkk..
now the line passing thru both the sources n meeting circle is having phase diff 99.4 λ n prependicular wali oλ ab to iske baad counting aani chahiye
1hey eureka. use tapan's formula. that's apt one. :)
first of all consider one fourth of the circle, θε[0,Π/2]
for maximum intensity points nλ/d = sin-1(Π/2) = 1 => n = d/λ . to get total no. of points for the entire circle, multiply n with 4, so max. intensity points will be 4n = 4(d/λ)
for minimum intensity points (2n'+1)λ/2d = 1 => n' = (2d-λ)/2λ. for the entire circle, min. intensity points will be 4n' = 2(2d/λ - 1)
answers will be
1 - p,r ; 2 - p,r ; 3 - q ; 4 - s.
21I forgot to mentipon one thing .........
In case OF MAXIMA dont forgot to remove cases that overlap
@Race : formula U used mein kuch subtraction chahiye........ THINK!!!
1@tapan, i got ur point. but mujhe jo answers aa rahe hain, they are in decimal points, so overlapping ka chance nahi hain. except for that d=100λ wala.
for that i think no. of Imax. points 396 aayenge and no. of Imin. points will be 398. !
but no answer matches for 3rd one then. !!
correct me if i am wrong anywhere. !
1waise jus find the phase diff from the path diff..
n calculate..
did u give all the info?
24column 1 is distance b/w both sources d
and column 2 is total no. of points of max. intensity or min. intensity on periphery of circle shown.
21USE THIS : sin(θ) = nλ/d ............. THIS REALATION IS FOR MAXIMA
nnow put range of SIN-1(....)