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Why not 7?Actually the answer is C. 6
just use dsin@=n*wavelength with @=90 and find n, you will get around 3 for XY and same for YZ
so in total its 6
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Why not 7?Actually the answer is C. 6
just use dsin@=n*wavelength with @=90 and find n, you will get around 3 for XY and same for YZ
so in total its 6
Why not 7?
use the equation nλ=dSinθ where n(number of order), λ(wave lengeth), d(distance between each grating) and θ being the angle.
Haha - That's a relief. I got mega-confused just then.yeah dude you're right. im mistaken
Haha - That's a relief. I got mega-confused just then.
Haha no harm done.lol sorry for that. Good luck for the exam today hope you do very well
answer is D, because only points within the overlap are in phase and so A and C are wrong. also B is wrong because stationary waves have varying amplitudes.
answer is D, because only points within the overlap are in phase and so A and C are wrong. also B is wrong because stationary waves have varying amplitudes.
27 shd b A.
thanks
so u mean to say from N1 TO N2 point ass in phase then from n2 to n3 is it?
27 shd b A.
No. 1. Nodes are always formed at closed ends so discard D.
No.2 In one loop, ie between two nodes, particles are in phase. WITHIN one segment. (Or loop)
So particles r in phase, one direction, in bottom segment.
Particles are in phase in second (top) segment. BUT particles in bottom and top segments ARE NOT in phase with EACH other.
I hope u get what I mean.
Amplitude of particles equidistant from a node is same.
So A.
Ur welcome.yes, i understood. thank you.
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