# How to find slit separation?

## Homework Statement

The green line of gaseous mercury at 546nm fals on a double-slit apparatus.
If the fifth dark fringe is at 0.150 degree from the centerline, what is the slit separaton?

## Homework Equations

d sin (theta) = m* lambda

## The Attempt at a Solution

ok i used the above equatiob to find the d, slit separation
as d=m*lambda/sin(theta)

but im getting the wrong answer....and im not able to figure out the problem....plez
som1 help.....!!

tiny-tim
Homework Helper
Hi patelpalak! (have a lambda: λ and a theta: θ )

Show us your full calculations, and then we can see what went wrong, and we'll know how to help! ok here's what i did

d sin (theta) =m*lamdba

therefore, d=m*lambda/ sin(theta)

m=5, since its fifth fringe
lambda= 546*10^-9
theta = 0.150 degrees

so, d=(5)(546*10^-9)/sin(0.150)

and i get 0.001042 meters
but it gives me wrong answer....!!!!
thats all i did ...

tiny-tim
Homework Helper
Hi patelpalak! (whatever happened to that λ and θ I gave you? )
ok here's what i did

d sin (theta) =m*lamdba

theta = 0.150 degrees

so, d=(5)(546*10^-9)/sin(0.150)

erm … θ has to be in radians! Hmmm.... If it is a dark fringe, isn't the equation something like:

dsinθ = (m+0.5)λ

?

Also, if my memory serves me correctly, you need to be careful with the order of your minimum. The first minimum has an m value of 0.

Last edited:
ok thanx all i got the answer....
appriciate your help....thanx once again!! 