Our disscusssion of the techniques for determining cosntructive and destructive by reflection from a thin film in air has been confused to rays striking the film a nearly the normal incidence.Assume that a ray is incident at an angle of `45^(@)` (relative to the normal ) on a film with an index of refraction of `sqrt(2)`.Calculate the minimum thickness fro constructive interference for issodium light of wavelength `600nm`.

Answer

`122.47nm`

Explanation

From snell's law
`1xxsin45^(@)=sqrt(2) sinr`
`sinr=(1)/(2)`
`r=30^(@)` Path difference=`2mutcosr-(lambda)/(2)…..(1)`
For constructive interference in reflection path diff. `nlambda …(2)`
From(1)&(2)
`2mut cosr-(lambda)/(2)=nlambda rArr 2mut cosr=(2n+1)(lambda)/(2)`
`t=((2n+1)lambda)/(4mucosr) rArr t_(min)=(lambda)/(4mucosr)`
`t_(min)=(600nm)/(4xxsqrt(2)xxcos30^(@)) rArr t_(min)=122.47nm`