Question

what is the least non zero thickness of a piece of glass with n=1.5, for which destructive interference of rod (650 nm) light in air can take place by reflection?

Answer 1

108.33 nm

Step-by-step explanation:

For destructive interference the thickness
`2t=(m+(1)/(2))(\lambda )/(n)` , for minimum m=0

So
`2t=(\lambda )/(2n)`

`t=(\lambda )/(4n)`

Here
`\lambda` is wavelength and n is order of index

So
`=(\lambda )/(4n)=(650)/(4* 1.5)=108.333nm`

So the least non zero thickness for destructive interference is 108.33 nm

Answer 2

t = 108.33 nm

Step-by-step explanation:

we know that for destructive interference we have following relation

`2t =(m+(1)/(2)) (\lambda)/(n)`

for minimum thickness m =0

therefore we have

`t = (lambda)/(4n)`

where
`\lambda = 650 nm`

refrective index n = 1.5

putting all value to get required value of thickness

`t = (650)/(4*1.5)`

t = 108.33 nm