Question

A double slit that is illuminated with coherent light of wavelength 644 nm produces a pattern of bright and dark fringes on a screen 6.00 cm from the slits. If the slits are 2783 nm apart, what is the distance on the screen between the 4th and the 2nd bright fringes on one side of the central maximum?
A)17.6 cm
B) 23.0 cm
C) 3.13 cm
D)14.7 cm
E)11.5 cm

Answer 1

Answer:

$2.77$ cm

Explanation:

$d$ = separation between the slits = 2783 x 10⁻⁹ m

$\lambda$ = wavelength of coherent light = 644 nm = 644 x 10⁻⁹ m

$D$ = Distance of the screen = 6 cm = 0.06 m

$y_{n}$ = Position of nth bright fringe

Position of nth bright fringe is given as

$y_{n} = \frac{nD\lambda }{d}$  

for n = 2

$y_{2} = \frac{nD\lambda }{d}$  

$y_{2} = \frac{(2)(0.06)(644\times10^{-9}))}{2783\times10^{-9}}$

$y_{2} = 0.0278$ m

for n = 4

$y_{4} = \frac{nD\lambda }{d}$  

$y_{4} = \frac{(4)(0.06)(644\times10^{-9}))}{2783\times10^{-9}}$

$y_{4} = 0.0555$ m

Distance between 4th and 2nd bright fringes is given as

$w = y_{4} - y_{2} = 0.0555 - 0.0278 = 0.0277 m$

$w = 2.77$ cm