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Qmechanic ♦
Qmechanic ♦
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I have a problem with interferometer fringe shift. I am learning it from my physics book. In the book, it is written, "As the L1 (Displacement of moveable mirror) is changed, the pattern of interference fringes is observed to shift. If M1 (Moveable Mirror) is displaced through distance equal to $\lambda /2$, a path difference of double of this displacement is produced, i.e equal to $\lambda$."

"As the L1 (Displacement of moveable mirror) is changed, the pattern of interference fringes is observed to shift. If M1 (Moveable Mirror) is displaced through distance equal to $\lambda /2$, a path difference of double of this displacement is produced, i.e equal to $\lambda$."

I want to know why, the path difference generated is double of displacement of mirror? Is it just an assumption? And tell me about that fringe shift.

I have a problem with interferometer fringe shift. I am learning it from my physics book. In the book, it is written, "As the L1 (Displacement of moveable mirror) is changed, the pattern of interference fringes is observed to shift. If M1 (Moveable Mirror) is displaced through distance equal to $\lambda /2$, a path difference of double of this displacement is produced, i.e equal to $\lambda$."

I want to know why, the path difference generated is double of displacement of mirror? Is it just an assumption? And tell me about that fringe shift.

I have a problem with interferometer fringe shift. I am learning it from my physics book. In the book, it is written,

"As the L1 (Displacement of moveable mirror) is changed, the pattern of interference fringes is observed to shift. If M1 (Moveable Mirror) is displaced through distance equal to $\lambda /2$, a path difference of double of this displacement is produced, i.e equal to $\lambda$."

I want to know why, the path difference generated is double of displacement of mirror? Is it just an assumption? And tell me about that fringe shift.

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dmckee --- ex-moderator kitten
dmckee --- ex-moderator kitten
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I have a problem with interferometer fringe shift. I am learning it from my physics book. In the book, it is written, "As the L1  (Displacement of moveable mirror) is changed, the pattern of interference fringes is observed to shift. If M1  (Moveable Mirror) is displaced through distance equal to [ lambda ] /2 $\lambda /2$, a path difference of double of this displacement is produced, i.e equal to [ lambda]$\lambda$."

I want to know why, the path difference generated is double of displacement of mirror? Is it just an assumption? And tell me about that fringe shift.

I have a problem with interferometer fringe shift. I am learning it from my physics book. In the book, it is written, "As the L1(Displacement of moveable mirror) is changed, the pattern of interference fringes is observed to shift. If M1(Moveable Mirror) is displaced through distance equal to [ lambda ] /2 , a path difference of double of this displacement is produced, i.e equal to [ lambda]."

I want to know why, the path difference generated is double of displacement of mirror? Is it just an assumption? And tell me about that fringe shift.

I have a problem with interferometer fringe shift. I am learning it from my physics book. In the book, it is written, "As the L1  (Displacement of moveable mirror) is changed, the pattern of interference fringes is observed to shift. If M1  (Moveable Mirror) is displaced through distance equal to $\lambda /2$, a path difference of double of this displacement is produced, i.e equal to $\lambda$."

I want to know why, the path difference generated is double of displacement of mirror? Is it just an assumption? And tell me about that fringe shift.

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Haris Ahmad
Haris Ahmad
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Michleson Interferometer fringe shift

I have a problem with interferometer fringe shift. I am learning it from my physics book. In the book, it is written, "As the L1(Displacement of moveable mirror) is changed, the pattern of interference fringes is observed to shift. If M1(Moveable Mirror) is displaced through distance equal to [ lambda ] /2 , a path difference of double of this displacement is produced, i.e equal to [ lambda]."

I want to know why, the path difference generated is double of displacement of mirror? Is it just an assumption? And tell me about that fringe shift.