Question

Wave Optics - Double Slit interference A helium neon laser (LaTeX: \lambda=633\:nmλ = 633 n m) illuminates a double slit that is 2 m from a screen. The light transmitted from the slits produces an interference pattern with a 3.2 mm spacing of successive bright fringes.

1. What is the spacing of the slits?
2. What frequency of laser light is needed to generate a 4.0 mm spacing of successive bright fringes?
3. How will the bright fringe spacing change if the laser is moved farther from the slits, the wavelength is halved, the distance from slit to screen is increased by a factor of 4 and the slit separation is doubled? Pick from list below

• No change
• Larger
• Smaller
• Different color

Answer 1

1. The fringe width , b is given by the formula:

b = kD/d ......eqn 1

where k is the wavelength of the incident radiation; D is the distance of the screen from the slits and d is the seperation between two slits

Here, wavelength k of incident light i.e laser =633 nm =633*10-9 m

Distance of screen from the slits, D = 2 m

and spacing between two bright fringes i.e fringe width b =3.2 mm = 3.2*10-3 m (1 mm = 10-3 m)

Using values of D, k and b in equation 1, we get

3.2*10-3 =633*10-9*2/d

or d = 633*10-9*2/3.2*10-3

d = 395.625*10-6 m

= 0.395*10-3 m

d = 0.39 mm =0.4 mm (approximately)

Hence, the seperation of slit is 0.4 mm

2. Required spacing between the successive bright fringes, b = 4 mm = 4*10-3 m

We would first calculate wavelength and then the frequency of required incident light;

d = 0.39mm = 0.39*10-3 m(calculated above)

D = 2 m

Again using the equation 1, and putting the values for D, d and b; we get

4*10-3 = k*2/0.39*10-3

or k = 4*10-3*0.39*10-3/2

k = 0.78*10-6 m

This gives the required wavelength of incident light for getting the fringe width as 4 mm

Now, since laser light is the form of electromagnetic radiation, it traveles with a speed equal to velocity of light c(=3*108 m/s) in the vacuum, hence

c = vk

where v is the frequency of the light;

i.e Using this formula; we can calculate the frequency corresponding the wavelength k of the laser light calculated above to obtain a fringe width of 4 mm

Hence, v = c/k

= 3*108/0.78*10-6

v = 3.846*1015 Hz ........Required frequency of the incident light

3.Currently we are doing this question considering that all the changes are to be done in a single go, if they are to be done individually, kindly revert back, we would be happy to help you with the accurate responses :)

We know fringe width, b = kD/d

Since, the formula doesnt involve the distance between the source of light and the slits, hence it would make no change in b.

Secondly, when wavelength k is halved, i.e k' = k/2; distance btw source and slits is increased i.e D' = 4D and slit seperation is halved; i.e d' = 2d; the new fringe width would be;

b' = k'D'/d'

= (k/2)*(4D)/2d

= 2kD/2d

b' = kD/d

Compare this eqn with eqn 1;

b' = b

i.e on making all these changes simulatenously; the fringe width remains same; i.e there is no change in the seperation between two bright fringes