Need solutions please. #24 answer is b: 133nm #25 answer is c 24) L1 = L2 gives

Question

Need solutions please.

#24 answer is b: 133nm

#25 answer is c

24) L1 = L2 gives the full 200 W coming to the detector. What is the minimum increase in L2 required so that only 100 W goes to the detector?

25) Assume that L2 has been increased by the amount calculated in the previous question, so 100 W is reaching the detector. The LIGO interferometer arms are kept under high vacuum. If a small amount of air now leaks into the system (filling both arms equally), what will initially happen to the power on the detector? Neglect any absorption of the light by the air.

The LIGO interferometer is used to detect the very very small compression and expansion of space as a gravity wave passes by. I is essentially a Michelson interferometer, each of whose arms L and L2 is 4 km long. The input is a 200-W laser at a wavelength of 1064 nm. Assume that at the start the interferometer arms are adjusted so that all the light exits to the detector shown. Basic Michelson Interferometer LoserD Beam Splitter 24) L L2 gives the full 200 W coming to the detector. What is the minimum increase in L2 required so that only 100 W goes to the detector? a. 266 nm b. 133 nm c. 532 nm 25) Assume that L2 has been increased by the amount calculated in the previous question, so 100 W is reaching the detector. The LIGO interferometer arms are kept under high vacuum. If a small amount of air now leaks into the system (filling both arms equally), what will initially happen to the power on the detector? Neglect any absorption of the light by the air. a. The power will increase slightly. b. The power will stay the same. c. The power will decrease slightly.

Explanation / Answer

#24) SO as the power to the detector falls by half, that means that the amplitude of the intering waves from both arms is not constructively interfaring

for the light wave equaiton of the form y = Asin(wt)
constructive interference gives amplitude 2A, so power is proportional to (2A)^2 = 4A^2
when power falls by half, means the resultant of the interference is an amplitude of sqroot(4A^2/2) = A*sqroot(2)

so for a phase shift of p
Asin(wt) + Asin(wt + p) = A*sqroot(2)*sin(wt + phi)
Sin A + Sin B = 2sin([A + B]/2)cos([A - B]/2)

so, 2A[sin(wt + p/2)cos(p/2)] = A*sqroot(2)sin(wt + phi)
hence
2A*cos(p/2) = A*sqroot(2)
cos(p/2) = 1/sqroot(2) =
p/2 = pi/4
p = pi/2

but 2*pi phase shift will correspond to a path difference of lambda
hence pi/2 phase shift means, path difference of lambda/4

to achieve this path difference, the mirror has to be moved half this distance, = lambda/8
but lambda = 1064 nm
so distance moved, d = 1064*10^-9/8 = 133 nm
so option b.

#25) when air seeps in the experiment chamber, the permittivity of free space changes, changin the amplitudes of electric and magnetic foelds and hence the power detection in the detector decrerases slightly

so option c.

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