SOLUTION The first dark fringes that flank the central bright fringe correspond

ID: 1437714 • Letter: S

Question

SOLUTION

The first dark fringes that flank the central bright fringe correspond to m= ±1.

Relate the position of the fringe to the tangent function.

tan = y1/L

Because is very small, we can use the approximation sin tan and then solve for y1.

sin tan y1/ L

Compute the distance between the positive and negative first—order maxima, which is the width w of the central maximum.

PRACTICE IT

Determine the width of the first-order bright fringe, when the apparatus is in air.
w1 =

sin = ± = ± 5.80 10-7 m = ±1.93 10-3 a 3.00 10-4 m

for m th dark fringe.

y = (m + 0.5)*lambda*D/d

y1 = (1 + 0.5) * 5.74 x 10^2 x 10^-9 x 1.92 / (0.284 x 10^-3)

y1 = 5.82 x 10^-3 m = 5.82 mm ........Ans

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width of central fringe will be = 2y1 = 2 x 5.82 = 11.6 mm

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