Question

) when light of 625 nm falls on a double slit whose spacing is 2.95×10-6 m? At most, how many bright fringes can be formed on each side of the central bright fringe (not counting the central bright fringe) when light of 625 nm falls on a double slit whose spacing is 2.95×10-6 m? A. 2 B. 3 C 4 D. 5 E. 6

Answer 1

To solve this problem it is necessary to apply the concepts related to the principle of overlap and constructive interference.

By definition we know that

$dsin\theta = n\lambda$

Where,

d = Distance between slits

n = Number of fringes (or number of repetition of the spectrum)

$\lambda$ = Wavelength

Our values are given as

$d = 6.95*10^{-6}m$

$\theta = 90\° \rightarrow$ The angle is 90 degrees because the angle is the furthest angle the light can rotate from out of the slits.

$\lambda = 625*10^{-9}m$

Replacing we have that

dsin\theta = n\lambda

$(2.95*10^{-6})sin(90) = n(625*10^{-9})$

n = \frac{(2.95*10^{-6})}{(625*10^{-9})}

n = 4.72

Therefore 4 complete bright fringes are formed and the number of bright fringes without including the central bright fringe is 3. The correct answer is B.