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In a double slit experiment, the distance between the slits is 5.0 mm and the slits are 1.0 m from the screen. Two interference patterns can be seen on the screen one due to light with wavelength 480 nm, and the other due to light with wavelength 600 nm. What is the separation on the screen between the third order bright fringes of the two interference patterns?

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0.072 mm

0.098 mm

0.088 mm

0.568 mm

answer is A.

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Detailed Solution

Distance:

$= \frac{3 λ_{2} D}{d} - \frac{3 λ_{1} D}{d} = \frac{3 (λ_{2} - λ_{1}) D}{d} = \frac{3 \times (600 - 400) \times 10^{- 9} \times 1.0}{5.0 \times 10^{- 3}} = 7.2 \times 10^{- 5} m = 0.072 m m$

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