14. Light with a wavelength of 646 nm passes through two slits and forms an interference pattern on a screen 8.75 m away. The distance between the central bright fringe and the second-order bright fringe is 5.16 cm. What will be the distance between the central bright fringe and the third-order minimum! A) 0.13 m B) 0.18 m C) 0.24 m D) 0.34 m E) 0.47 m

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EXPERT ANSWER

According to question, light has wavelength, \lambda = 646 nm = 646 * 10-9 m.

distance of screen from slits, D = 8.75 m

Distance between central bright fringe and second order bright fringe is, y = 5.16 cm = 5.16 * 10-2 m. i.e for n = 2

Since, for maxima

—Yи
= R

Where d is distance between slits and n is order of maxima

Therefore, on rearranging

D
d=n
Y
d = 2 * 646 * 10-9
*
8.75
5.16 * 10-2

d=2.19 * 10 m —–equation 1

For minima, distance between central bright fringe and nth order minima (or dark fringe) is

y=(2n-1)\lambda \frac{D}{2d}

putting all given values and calculated value(equation 1) in above equation we get

8.75
y = (2n – 1) * 646 * 10-9*
2 * 2.19 * 10-4

Therefore for n = 3

y = (2*3-1) * 646 * 10-9*
8.75
2 * 2.19 * 10-4
у —
0.0645m = 6.45cm

Therefore distance between the central bright fringe and third order minimum is 6.45 cm or 0.0645 m.