Question

For a double slit experiment, what is the smallest separation (in nm) between two slits that will produce a second-order maximum for 596 nm visible light? You need to round your answer to the nearest integer, do not include unit.

Answer 1

We know that for constructive interference:

m* = d*sin

for 2nd order maximum, m = 2

= wavelength of visible light = 596 nm

for smallest separation distance, sin should be maximum, which is

sin = 1, So

d = separation distance between two slits = m*/sin

d = 2*596/1

d = 1192 nm (do not use the units 'nm' as mentioned)

If you need answer in 'm' units then use d = 1.192 m

Let me know if you've any query.