The intensity on the screen at a certain point in a double-slit interference patteren is 25.0% of the maximum value. (a) What minimum phase difference (in radians) between the sources produces this result? (b) Express this phase difference as a path difference for 486.1 nm light.

The intensity on the screen at a certain point in a double-slit interf

1.	The intensity on the screen at a certain point in a double-slit interference patteren is 25.0% of the maximum value. (a) What minimum phase difference (in radians) between the sources produces this result? (b) Express this phase difference as a path difference for 486.1 nm light.
Answer» (a) `(I)/(I_(max)) = cos^(2) ((phi)/(2))` Therefore, `phi = 2 cos^(-1) sqrt((I)/(I_(max))) = 2 cos^(-1) sqrt 0.25` `2 cos^(-1) ((1)/(2)) = 2 xx (pi)/(3) = (2pi)/(3) rad` (b) `delta = (lambda phi)/(2 pi) = ((486 nm)((2)/(3) pi "rad"))/(2 pi) = 162`

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The intensity on the screen at a certain point in a double-slit interference patteren is 25.0% of the maximum value. (a) What minimum phase difference (in radians) between the sources produces this result? (b) Express this phase difference as a path difference for 486.1 nm light.

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