# Angular Width Of Central Diffraction

1. Jun 1, 2016

### 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?

2. Jun 1, 2016

### blue_leaf77

Are those numbers in the same unit?

3. Jun 1, 2016

### Chris18

λ=nm and d= mm

4. Jun 1, 2016

### TSny

Note 1 mm ≠ 10-6 m.

5. Jun 1, 2016

### Chris18

Ok so I must have 10^-7 ?

6. Jun 1, 2016

### blue_leaf77

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 $\lambda$.