How to find slit separation?

[patelpalak]

1. The problem statement, all variables and given/known data

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?


2. Relevant equations

d sin (theta) = m* lambda

3. 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]

Hi patelpalak! :smile:

(have a lambda: λ and a theta: θ :wink:)

Show us your full calculations, and then we can see what went wrong, and we'll know how to help! :smile:

[patelpalak]

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]

Hi patelpalak! :smile:

(whatever happened to that λ and θ I gave you? :confused:)
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 :redface: … θ has to be in radians! :wink:

[tyco05]

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.

[patelpalak]

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