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Consider a ray of light incident from air onto a slab of glass (refractive index n) of width d, at an angle theta. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is |
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Answer» `(2pind)/(lamda)(1-(sin^(2)theta)/(n^(2)))^(-1//2)+pi`  Situation given in the statement is depicted in above figure. Suppose light ray `vecPQ` is made incident at point Q on surface t=0. Here refracted ray `vecQS` gets reflected from point S after sometime, say after time t. Now, if SPEED of light in glass is v then, `v=(QS)/(t)` `:.t=(QS)/(v)` `:.t=(((d)/(cosr)))/((c)/(n))""(":.cosr=(d)/(QS)andn=(c)/(v))` `:.t=(nd)/(cxxcosr)""....(1)` Now, APPLYING Snell.s law at point Q, `sintheta=(n)sinr` `:.n=(sintheta)/(sinr)` `:.sinr=(sintheta)/(n)` `:.sin^(2)r=(sin^(2)theta)/(n^(2))` `:.1-cos^(2)r=(sin^(2)theta)/(n^(2))` `:.cosr=(1-(sin^(2)theta)/(n^(2)))^(1/2)""....(2)` From equation (1) and (2), `t=(nd)/(cxx(1-(sin^(2)theta)/(n^(2)))^(1/2))` `:.t=(nd)/(c)xx(1-(sin^(2)theta)/(n^(2)))^(-1/(2))""......(3)` Now, at the end of time t, if phase of ray `vecST` is `phi_(1)` and phase of ray `vecQR` is `phi_(2)` then phase difference between them is `Deltaphi=phi_(2)-phi_(1)=omegat` `:.Deltaphi=(kc)t""("":.c=(omega)/(k)=("angular frequency")/("wave vector"))` But since ray `vecQR` gets reflected from the surface of denser MEDIUM, net path difference would be `Deltaphi.=Deltaphi+pi` `:.Deltaphi.=(kc)t+pi` `:.Deltaphi.=((2pi)/(lamda)c)((nd)/(c))(1-(sin^(2)theta)/(n^(2)))^(-1/2)+pi` ( `:.k=(2pi)/(lamda)` and from equation (3)) `:.Deltaphi.=(2pind)/(lamda)(1-(sin^(2)theta)/(n^(2)))^(-1/2)+pi` |
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