The intensity on the screen at a certain point in a double...

The intensity on the screen at a certain point in a double slit interference pattern is $64.0 \%$ of the maximum value.
(a) What minimum phase difference (in radians) between sources produces this result? (b) Express this phase difference as a path difference for 486.1 -nm light.

Key Concepts

Double-Slit Interference

Double-slit interference is a phenomenon where waves (often light) from two coherent sources superpose and produce a pattern of alternating bright and dark regions on a screen. This effect arises because the waves interfere constructively (in phase) or destructively (out of phase), leading to variations in intensity at different points on the screen.

Interference Intensity Formula

The intensity of the superposed waves in interference patterns is typically described by a cosine squared relation, where the intensity is proportional to cos²(?/2), with ? representing the phase difference between the waves. This formula allows one to derive the phase difference corresponding to a specific intensity fraction of the maximum value.

Phase Difference

Phase difference is a measure of the displacement between the peaks (or other corresponding points) of two waves. It is an angular measure, typically expressed in radians, and it determines whether the waves will interfere constructively or destructively when they combine.

Path Difference

Path difference is the difference in the distance traveled by two waves from their respective sources to a point of interest. This physical separation translates into a phase difference and is related to wavelength by the equation ? = (2?/?) × (path difference), allowing conversion between a phase difference and a corresponding physical distance.

Transcript

00:01 Hi, in young's double slit experiment, the resultant amplitude of the waves coming out of two silutes on the screen is given as a square is equal to a1 square plus a 2 square plus 2 a 1, a 2, cause 5.

00:25 Where a1 and a2 are the amplitudes of the waves coming from the two slits.

00:52 And phi is the phase difference between these two waves when they reach at screen.

01:25 If we convert this expression in the form of intensity, the resultant intensity at the screen is given as i is equal to i1 plus i2 plus 2 root i 1 into i2 cos 5 using the relation between the intensity and amplitude which says intensity depends directly on the square of amplitude.

01:55 So if the same intensity comes out of the two slits, means if we consider i1 is equal to i2, let it be x, then the resultant intensity at the screen will be given by x plus x plus 2, square root of x into x into cos phi, or it becomes 2x plus 2x plus 2x cos phi.

02:28 For maximum, maximum intensity at the screen means for imex, this phi, phase difference between the waves should be 0 degree.

02:45 So this cos phi will come out to be 1.

02:47 Hence, expression for maximum intensity becomes 2x plus 2x into 1 means 4x.

02:57 Now in the problem, the resultant intensity at a point is given as 64 % of the maximum intensity means this is 64 by 100 multiplied by this 4x or it comes out to be 2 .56 times of x.

03:23 So putting this value here in this expression in equation number one we can say.

03:30 So using equation one here which says i is equal to to 2x plus 2x cos 5.

03:43 Here for i this is 2 .56 x is equal to 2x plus 2x cos 5...

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