Angular Width Of Central Diffraction

[Chris18]

350nm falls on a single slit of width of 0.20mm. What is the angular width of the central diffraction peak?
I think that the width should be equal to 2Θ, where Θ=arcsin(m*λ/d)... m=1 and we have λ=350*10^-9 and d= 0.20*10^-6...but when i do the calculations I get 1.75 and the arcsin is a maths error...Could anyone help me please?

 

[blue_leaf77]

λ=350*10^-9 and d= 0.20*10^-6

Are those numbers in the same unit?

 

[Chris18]

λ=nm and d= mm

 

[TSny]

Note 1 mm ≠ 10^-6 m.

 

[Chris18]

Ok so I must have 10^-7 ?

 

[blue_leaf77]

d= 0.20*10^-6

You should have written the unit as well, if you write it correctly it will be $d= 0.20 \times 10^{-6}$ km. If you use km for d then you must also use km for $\lambda$.