The change in internal energy that accompanies the transfer of heat, q, or work, w, into or out of a system can be calculated using the following equation: Note the value of heat and work as they are transferred into or out of a system.
W= 18*10^-3m
N= 18*710 ( because you have 710 slits per mm ) = 12780
lambda = 506*10^-9
The first step is to workout d.
d = w/N
d =18*10^-3m/12780
d =1.41*10^-6m
The second step is to work out the maximum number of m.
Since sin (theta) = less than 1, then m*lambda/d = less then 1,
therefore m= less than d/lambda
( I know thats confusing but trust me )
so m is less than 1.41*10^-6m/506*10^-9
= 2.7
Therefore use m = 2
Lastly put it all into the formula
dsin (theta) = m*lambda so:
theta = sin^-1(m*lambda/d)
theta = sin^-1(2*506*10^-9/1.41*10^-6m)
theta = 45.95 degrees or 46 degrees
Because of the different speeds..
