In the figure below, a light wave along ray r_1 reflects once from a mirror and

Question

In the figure below, a light wave along ray r_1 reflects once from a mirror and a light wave along ray r_2 reflects twice from that same mirror and once from a tiny mirror at distance L from the bigger mirror. (Neglect the slight tilt of the rays.) The waves have wavelength 610 nm and are initially in phase. What is the smallest value of L that puts the final light waves exactly out of phase? With the tiny mirror initially at that value of L, how far must it be moved away from the bigger mirror to again put the final waves out of phase?

Explanation / Answer

a)

path differrnece = 2L

phase difference = pi for out of phase

2pi/610*10^-9 * 2L = pi

so L = 152.5 nm

b)

let it shift by x for second reflection ... then path difference = 2*L + 2*(L+x) = 4L + 2x = 610 + 2x

2pi/610*10^-9 * (610+2x) = 3pi

610nm+2x = 915 nm

x =152.5 nm

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