Question

Light of wavelength 400nm is incident on two slits separated by 1000mm. The interference pattern from the slits is observed from a satellite orbiting 0.4Mm above the Earth. The distance between interference maxima as detected at the satellite is

A. 0.16Mm.
B. 0.16km.
C. 0.16m.
D. 0.16mm.

Answer 1

C) 0.16 m

Step-by-step explanation:

Distance between two consecutive maximum is given as

`\beta = (\lambda L)/(d)`

here we know that

L = distance of the screen = 0.4 Mm

d = distance between two slits = 1000 mm

`\lambda` = 400 nm[/tex]

so we know that

`\beta = (400* 10^(-9) (0.4 * 10^6))/(1)`

`\beta = 0.16 m`

Answer 2

We are given:

λ, Light with a wavelength of 400nm
w Separation of slits = 1000mm
L, Distance from the orbit = 0.4Mm


λ = zw/(mL)

400x10^-9 = z* 1000x10^-3 / 0.4x10^6 m
Solve for z, this is the distance between interference maxima as detected at the satellite