# 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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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

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

Therefore, on rearranging

$\Rightarrow d=2.19*10^{-4} m$ —–equation 1

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

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

Therefore for n = 3

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