Question

A single slit diffraction experiment performed with an argon laser of wavelength 454.6 nm produces a pattern on a screen with dark fringes (minima) separated by 10 mm. If we switch to a helium laser of wavelength 632.8 nm, without changing anything else in the setup, what will be the new separation between dark fringes? Select one:

a. 7.2 mm

b. 6.1 mm

c. 13.9 mm

d. 17.3 mm

e. 15.2 mm

Answer 1

To develop this problem it is necessary to apply the concepts related to Interference and the Two slit experiment.

Young's equation defines the separation between fringes this phenomenon as,

$d = \frac{\lambda D}{a}$

Where,

$\lambda =$ Wavelength

d = Separation between fringes

a =  Slit width

D = Distance between the slits and screen

Then for the two case we have

$d_1 = \frac{\lambda D}{a_1}$

$d_2 = \frac{\lambda D}{a_2}$

Calculating the new separation between the fringes would be

$\frac{d_2}{d_1} = \frac{\frac{\lambda D}{a_1}}{\frac{\lambda D}{a_2}}$

$\frac{d_2}{d_1} = \frac{\lambda_1}{\lambda_2}$

$d_2= \frac{\lambda_1}{\lambda_2} d_1$

We have then,

$d_2 = \frac{632.9*10^{-8}}{454.6*10^{-9}} 10*10^{-3}$

$d_2 = 13.9mm$

Therefore the correct answer is c.