Question

A double-slit interference pattern is created by two narrow slits spaced 0.15 mm apart. The distance between the first and the fifth minimum on a screen 80 cm behind the slits is 6.5 mm. What is the wavelength (in nm) of the light used in this experiment? Round off your answer to 1 decimal place (ex. 304.5 nm) and include the appropriate units

Answer 1

Distance between the slits d = 0.15mm 1.5×10-4 m .

Difference between minima fifth and first x5 - x1= 6.5mm

Screen distance D =80cm = 0.8m

Now condition at point of minima is

d*x/D = (n - 1/2)λ

Where λ is wavelength of light.

Now for order n= 1

d*X1/D = (1-1/2)λ

So 1.5×10-4×(x1)/0.8 = λ/2

So X1 = 2666.7λ

Now for order n= 5

d*X1/D = (5-1/2)λ

So 1.5×10-4×(x2)/0.8 = 9λ/2

So X2 = 24000λ

Now this difference is already given

x5 - X1 = 6.5mm

So (24000- 2666.7)λ = 6.5×10-3m

So 21333.3λ = 6.5×10-3

λ = 304.7nm.

So wavelength of light is 304.7nm.