Question

If two slits are separated by .25cm and the slit to screen is 1m, what is the distance between consecutive maxima for light of wavelength 0.5um? Show steps

Answer 1

Solution: We know for interference pattern of with two slits, for bright fringes (maxima) we have the relation,

d*sinθ = m*λ

where d = distance between two slits,

θ = angular position of the bright fringe (maxima) with respect to the central axis (central maxima),

m = order of the bright fringe and

λ = wavelength of the light used.

We can write above equation as,

sinθ = m*λ/d

If θ is small, then using small angle approximation, we haven sinθ ~ θ, that is θ sinθ is nearly equal to θ when the angle are small. Thus above equation becomes,

θ = m*λ/d           -----------------------------------------------------------------(1)

Now in the diagram, consider triangle OAB; we can write

tanθ = Y_m/X

where Y_m = distance between the m^th order maxima and the center of the screen and

X = distance between the slits and the screen.

Again using small angle approximation, tanθ ~ θ. Thus above equation becomes,

θ = Y_m/X                               -----------------------------------------------------------------(2)

From equation (1) and (2), we have,

m*λ/d = Y_m/X

Y_m = X*m*λ/d

Consider the next maxima which is even farther away from the center of the screen,

Y_m+1 = X*(m+1)*λ/d

Thus the distance between two arbitrary consecutive maxima is given by

Y_m+1 – Y_m = (X*(m+1)*λ/d) – (X*m*λ/d)

Y_m+1 – Y_m = X*λ/d             --------------------------------------------------------------------(3)

We are given, slit separation d = 0.25 cm = 0.0025 m

slit to screen distance X = 1 m

Wavelength of light used λ = 0.5 μm = 0.5*10^-6 m

The distance between the consecutive maxima; Y_m+1 – Y_m = ? to be found.

Thus from equation (3), we have

Y_m+1 – Y_m = (1m)*( 0.5*10^-6m)/(0.0025m)

Y_m+1 – Y_m = 2*10^-4 m

Thus the distance between the two consecutive maxima is 2*10^-4 m.

(or 0.2 mm)