Light of wavelength 630 nm from a distant source is falling on a slit of width 0.5 mm. The diffraction pattern is observed on a screen 3.00 m away. What is the width of the central maximum?

### 1 Answer | Add Yours

general equatin:

sin(theta) = m`lambda` /a ; m = ...,-2,-1,0,1,2,...

central maximum ==> m = 1

so:

sin(theta) = y/L = (`lambda`)/a

==> y = L `lambda`/a = 3 * 630e-9/0.5e-3 = 0.00378 m

==> the width of the central maximum = 2*0.00378 = 0.00756 m = 7.56 mm

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes