Answer:
Explanation:
At both interface light is travelling from a medium of lower refractive index to medium of higher refractive index. So the phase difference of 180⁰ occurring at both interface will cancel out .
Net path difference = 2 μ t where μ is refractive index of thin layer and t is its thickness
For destructive interference
2 μ t = (2n + 1 )λ / 2 where λ is wavelength of light , n is any number
λ = 4 μ t / ( 2 n + 1 )
For largest wavelength , n = 0
λ = 4 μ t
= 4 x 1.22 x 138 nm
= 673.44 nm .
