Question

A screen containing two slits 0.100mm apart is 1.50m from a viewing screen. Light of wavelength 633nm falls on the slits. Approximately how far apart are the bright fringes on the viewing screen?

Answer 1

Answer:

distance = 9.495 mm

Step-by-step explanation:

Given data

slits distance = 0.100 mm

viewing screen distance = 1.50 m

wavelength =  633 nm

to find out

how far apart are the bright fringes on the viewing screen

solution

we use here double slit diffraction condition that is

d sin(θ) = m λ    .......................1

here m = 0, 1 , 2 , 3   .....

and here for θ small so

sin(θ) = tan(θ) = $\frac{x}{l}$

so x = l sin (θ)

so from equation 1

x = $\frac{m\lambda l}{d}$

and

x1 =  $\frac{\lambda l}{d}$     ...............2

x2 =  $\frac{2\lambda l}{d}$   ..............3

so distance x2 - x1  will be

distance = $\frac{\lambda l}{d}$

distance = $\frac{633*10^{-9}*1.5}{0.1*10^{-3}}$

distance = 9.495 mm