Calculating slit width by length difference of 1st and 2nd minima

[Unemployed]

Homework Statement

What is the slit width? Wavelength is 635 nm. Distance between screen and slit, 2.5 m. Distance on screen between first and second minima above the central maxima is delta 6 mm.

Homework Equations

D* sin theta * = m * lambda

tan theta = x/ distance between screen and slit

The Attempt at a Solution

I set lambda/sin theta m2-sin theta m1= 2*lambda/ sin theta m2

tan theta 2m = x + 6 *10^-3 m/2.5

I tried to plug it in, but i got stuck. What's the correct approach?

 

[rl.bhat]

Hi Unemployed, welcome to PF.
You have given the relevant equations. Now what is the condition for minimum on the screen? Write down two equations for x values.. One for first minimum and other for second minimum. Then find the difference between the x values. x2 - x1 value is given. Find D.

 

[Unemployed]

x1 = tan theta m1*2.5
x2 = (tan (theta m1 + theta m2) *2.5) + 6 *10^-3

Is theta for m2 half of what it is for m1?

Still puzzled.

 

[rl.bhat]

for small angle sin θ can be written as tan θ = x/D, where x os the distance between screen and slit. Put m = 1 for first minimum and m = 2 for second minimum. Find x1 and x2. Difference between x1 and x2 is given. Find slit width d using the formula x/D = m*λ/d