intesities...match the column.......

counitng matlab...woh kaise?[7] ...zaroor batana[2]

give the soln.....................
i am struggling with this for so many daysssss

is it multiple crrct n r u having the ans

in 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

samajh nahin aaya......[7][7]

hey 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.

yup...sure

method of the race n mine is same

@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 I_max. points 396 aayenge and no. of I_min. points will be 398. !

but no answer matches for 3rd one then. !!

correct me if i am wrong anywhere. !

i got 1-pr
2-ps
3-qs
4-s

nooo

points mean points on perphery of circle(imaginary ) drawn

!!!!! okies........

column 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.

achha no radius of the circle given ??

wat doest this mean?
398 points n 400 points etc...

USE THIS : sin(θ) = nλ/d ............. THIS REALATION IS FOR MAXIMA

nnow put range of SIN^-1(....)

tapan how does that help????????