Question

`ABC` is an isosceles right angled triangular refecting prism of average refractive index `mu`. When incident rays on face `AB` are paralel to face `BC`, then they emerges from Face `AC`, which are also parallel to Face `BC` as shown in figure-I. The prims capable to do so, known as Dove Prism. In figure II, the Dove Prism is used for dispersion of incident light containing red colour and violet colour only. The red colour and violet colour lights are separated (displaced) on screen by a distance `Deltal`. In reality, each ray of any colour has some width, whcih can be designated as `d`. It is clear that an observer candistinguish the red and violet rays that emerges from prism only if `Deltalged`. Otherwise the bundles of rays will overlap and mix.
[Given for Dove Prism in figure II: `mu_(R)=sqrt(5/2),Deltamu=mu_(V)-mu_(R)=0.02, sqrt(5)=2.25, EF1m` and `AB=4cm`]

Choose the INCORRECT option
A. As per figure-I, the average refractive index of Dove Prism may be greater than 1
B. As per figure I the average refractive index of Dove Prism may be greater than `sqrt(2)`
C. As per figure II ,the displacement `Deltal` depends upon average refractive index `mu` and length of `AB`
D. As per figure II, he displacement `Deltal` depends upon average refractive index `mu` and length of `EF`

Answer 1

Correct Answer - D
In figure -I, total internal reflection take place at face `AB`.
`/_EMD+/_MDE=/_BEM`
`implies beta(90-alpha)=45^(@)`
`impliessin(beta+45^(@))gt(1/(mu))`
`implies sin beta+cosbeta gt (sqrt(2))/(mu)`……….(1)

According to law of refraction at point `M`, we can write
`1xxsin(45^(@))=musin betaimpliessin betas=1/(musqrt(2))impliescosbeta=sqrt(1-1/(2mu^(2))`
Putting these value in equation (1), we have
`1/(musqrt(2))+sqrt(1- 1/(2mu^(2))) gt (sqrt(2))/(mu)implies sqrt(2mu^(2)-1) gt 1implies mu gt 1`
In figure II: Replace the Dove Prism, by glass cube of side as `AB`

`Deltal=(DeltaL)/(sqrt(2))`.............(2)
From geometry, we can write
`sinbeta=1/(mu_(R)sqrt(2))` and `sin (beta-Deltabeta)=1/((mu_(R)+Deltamu_(R))sqrt(2))`
`DeltaL=a tan beta-a tan (beta-Deltabeta)=1/(sqrt(2mu_(R)^(2)-1))-a/(sqrt(2(mu_(|r)+Deltamu)^(2)-1))`
`impliesDelta=a/(sqrt(2mu_(R)^(2)-1))(1-sqrt((2mu_(R)^(2)-1)/(2(mu_(R)+Deltamu)^(2)-1)))=a/(sqrt(mu_(R)^(2)-1))(1-sqrt((2mu_(R)^(2)-1)/(2(mu_(R)+Deltamu)^(2)-1)))`
`impliesDeltaL=a/(sqrt(2mu_(R)^(2)-1))(1-sqrt(2mu_(R)^(2)-1)/(2mu_(R)^(2)+4mu_(R)Deltamu+2(Deltamu)^(2)-1))`
Neglecting the value `(Deltamu)^(2)`, we have
`impliesDelta=a/(sqrt(2mu_(R)^(2)-1))(1-sqrt(1/(1+(4mu_(R)Deltamu)/(2mu_(R)^(2)-1))))-a/(sqrt(2m_(R)^(2)-1))(1-{1+(4mu_(R)Deltamu)/(2mu_(R)^(2)-1)}^(-1/2))`
Expand binomially, we have
`impliesDeltaL=(2amu_(R)Deltamu)/((2mu_(R)^(2)-1)^(3/2))`
Putting this value in equation (2), we have
`Deltal=(sqrt(2)amu_(R)Deltamu)/((2mu_(R)^(2)-1)^(3/2))`
Second method:
`DeltaL=(d/(dbeta)(a tan beta))xxDeltabeta`
`DeltaL=(asec^(2)beta)xxDeltabeta`
Also `sinbeta=1/(sqrt(2)mu_(R))`
`impliescos beta Deltabeta=1/(sqrt(2)mu_(R)^(2))Deltamu`
`impliesDeltaL=(asec^(2)betaxxDeltamu)/(sqrt(mu_(R)^(2)cosbeta`
`impliesDeltaL=(2amu_(R)Deltamu)/((2mu_(R)^(2)cosbeta)`
Third method:
`Deltal=(DeltaL)/(sqrt(2)), sin alpha = 1/ (mu_(r) sqrt(2)), sin beta=1/(mu_(v)sqrt(2))`
`DeltaL=atan alpha =a tan beta`
`=[1/(sqrt(2mu_(r)^(2)-1))-1/(sqrt(2mu_(v)^(2)-1))]`
`=a[1/2-1/(2sqrt(1+mu_(r)Deltamu))]`
`=a/2 [a-(1+mu_(r)Deltamu)^(-1//2)]`
`=a/2[1-1+1/2mu_(r)Deltamu]`
`=a/4mu_(r)Deltamu`
`Deltal=(DeltaL)/(sqrt(2))=a/(4sqrt(2)) mu_(r)Deltamu=(4cm)/(4sqrt(2))xx(sqrt(5))/(sqrt(2))xx2/100`
`=(sqrt(5))/100 cm = sqrt(5)xx10^(-4)m`
`=2.25xx10^(-4)m`
Since : `Deltal ge d`
`impliesd le 2.25xx 10^(-4)m`
`sqrt(2mu_(v)^(2)-1)=sqrt(2(mu_(r)+Deltamu)^(2)-1)`
`impliessqrt(4+4mu_(r)Deltamu)[(Deltamu)^(2)~~0]`
`=2sqrt(1+mu_(r)Deltamu)`