Question

A double-slit interference pattern is created by two narrow slits spaced 0.18 mm apart. The distance between the first and the fifth minimum on a screen 62 cm behind the slits is 5.9 mm. What is the wavelength of the light used in this experiment? (in nm)

Answer 1

The wavelength is
`\lambda = 4.28*10^(-4)mm`

Step-by-step explanation:

Generally wavelength is mathematically represented as

`\lambda = (D)/(L) * \Delta y`

Where
`\Delta y` is the difference between the minimums and its value is = 5.9mm

D is the slit spacing with a value = 0.18 mm

L is the distance between the slit and the screen with

value = 6.2cm =
`= 6.2 *10=620mm`

The distance first and the fifth minimum is usually = 4
`\lambda`

So the equation above becomes

`D * (\Delta y)/(L) = 4 \lambda`

Making wavelength the subject

`\lambda = (D * (\Delta y)/(L) )/(4)`

Substituting value

`\lambda = (0.18 * (5.9)/(620) )/(4)`

`\lambda = 4.28*10^(-4)mm`