ABC is an isosceles right angled triangular reflecting prism of average refractive index mu . When incident rays on face AB are parallel to face BC , then they emerges from Face AC . which are also parallel to Face BC as shown in figure-l. The prism capable to do so , known as Dove Prism. In figure-ll, 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 Af. In reality , each ray of any colour has some width, which can be designated as d. It is clear that an observer can distinguish the red and violet rays that emerges from prism only if Delta/varphige d .Otherwise the bundles of rays will overlap and mix. Q Find the maximum value of width d of bundle of rays that can be resolved at the outlet of Dove Prism as shown in the figure-ll.

ABC is an isosceles right angled triangular reflecting prism of averag

1.	ABC is an isosceles right angled triangular reflecting prism of average refractive index mu . When incident rays on face AB are parallel to face BC , then they emerges from Face AC . which are also parallel to Face BC as shown in figure-l. The prism capable to do so , known as Dove Prism. In figure-ll, 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 Af. In reality , each ray of any colour has some width, which can be designated as d. It is clear that an observer can distinguish the red and violet rays that emerges from prism only if Delta/varphige d .Otherwise the bundles of rays will overlap and mix. Q Find the maximum value of width d of bundle of rays that can be resolved at the outlet of Dove Prism as shown in the figure-ll.
Answer» `2.25xx10^(-3)`m `2.25xx10xx^(-4)`m `1.125xx10^(-3)` `1.125xx10^(-4)`m Solution :In figure-I, total internal reflection takes palces at face AB., `/_EMD + /_ MDE = /_ BEM` `beta + (90^@ - alpha) = 45^@ implies alpha = beta+ 45^@` `implies sin(beta + 45^@)> (1/mu)` `sin beta + cos beta gtsqrt(2)/mu`…..(1) According to law of refraction at POINT M, we can write `1xxsin45^@= musn beta implies sin beta = 1/(musqrt(2)) implies cos beta= sqrt(1 - 1/(2mu^(2)))` Putting these value in equation (1), we have `1/(musqrt(2)) + sqrt(1 - 1/(2mu^(2))) > sqrt(2)/mu implies sqrt(2mu^(2) -1) > implies mu gt 1` In figure-II : Replace the Dove Prism by glass CUBE of side as AB. `Delta L =(DeltaL)/sqrt(2)` ....(2) From geometry, we can write `sin beta = 1/ (mu_(r)sqrt(2))` and `sin(beta - Deltabeta)= 1/((mu_(R) + Deltamu_(R))sqrt(2))` `Delta L = atan beta - a tan (beta- Deltabeta) = a/sqrt(2mu_(g)^(2)-1) - a/sqrt(2(mu_(R) + Deltamu)^(2) -1)` `DeltaL = a/sqrt(2mu_(R)^(2) -1 )(1- sqrt((2mu_(R)^(2)-1)/(2mu_(g)^(2) + 4mu_(R)Deltamu + 2(Deltamu)^(2) -1)))` Neglecting the value `(Deltamu)^(2)`, we have, `implies DeltaL = a/(sqrt(2mu_(R)^(2)-1)) (1 -sqrt(1/(1+ (4mu_(R)Deltamu)/(2mu_(R)^(2)-1)))) = a/sqrt(2mu_(R)^(2) -1) ( 1-{1 + (4mu_(R)Deltamu)/(2mu_(R)^(2)-1)}^(1/2))` Expand binomically, we hav `+> Delta L = (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)(atanbeta))xxDeltabeta` `DeltaL=(asec^(2)beta)xxDeltabeta` Also `sin beta= 1/(sqrt(2)mu_(R))` `implies cos betaDeltabeta= 1/(sqrt(2)mu_(R)^(2)) Deltamu` `implies Delta L = (asec^(2)betaxxDeltamu)/(sqrt(mu_(R)^(2)cosbeta)` `DeltaL = (2amu_(R)Deltamu)/((2mu_(R)^(2)-1)^(3/2))`

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