Question

Blue light of wavelength 470 nm is used to illuminate a pair of narrow slits that are 0.020 mm apart and 1.60 m from a screen.

(a) What is the angular position of the second-order minimum (dark spot)?

(b) What is the distance on the screen between the central maximum and the second-order

minimum?

(c) The reason there is a dark spot at this location on the screen is because light from one slit has

to travel further than light from the other slit. What is this extra distance in terms of wavelengths

(or half wavelengths)?

(d) Circle one word from each underlined pair:

If you wanted a narrower interference pattern (i.e. you wanted to decrease the angle and distance

you found in parts (a) and (b)), you could use yellow / violet light instead of blue light, or you

could increase / decrease the distance between the slits.

Answer 1

(a) Given the wavelength of light () = 470nm = 470 x 10-9m

Seperation between the slits, d = 0.020mm = 0.020 x 10-3m

Distance from slit to screen, D = 1.60m

Number of order, n = 2

The angular position of the nthorder minima is given by,

So the angular position of the second order minima is

(b) Distance of the nthorder minima from centre is given by,

So the distance of the second order minima from centre is

(c) Path difference (extra distance) for nth order minima is

For the second order minima, the path difference (x) is

(d) From the equations of position and angular position of maxima and minima, it is clear that they are directly proportional to wavelength of the light used. So inorder to narrow the interference pattern, a light of low wavelength must be used. From yellow and violet, to get a narrow interference pattern, violet should be used because it has lower wavelength compared to yellow.

On the other hand distance between the slits (d) is inversely proportional to the position and angular position. So we should increase the seperation of slits inorder to narrow the interference pattern.