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Qn.(7)

Shift S of fringe pattern due to introduction of transparent plate in front of one slit is given by

S = D×t×(μ-1) / d   .....................(1)

Fringe width β = λ×D/d  .........................(2)

 where D is distance between slits and screen, t is thickness of plate, μ is refractive index of plate,

d is distance between plate and λ is the wavelength of light used.

 Shift in terms of fringe width S/β is obtained by dividing eqn.(1) by eqn.(2), 

 S/β= t×(μ-1) / λ = 2×10^-2/(600×10^-9)= (5/3)×10⁴ ............(3)

 if S/β is integer then shift gives bright fringe (maxima) at centre .

if S/β is of the form n+(1/2), where n is integer, then shift gives dark fringe (minima) at centre.

 Hence from eqn.(3)  the shift of fringe pattern in this particular case due to introduction of transparent plate

in front of one of the slit gives neither maxima nor minima

Qn.(8)

Let μ_i be the initial refractive index and μ_f is the refractive index after heating of plate.

 Difference of shift pattern on the screen = [ (  D×t×(μ_f-1)/d )  -  ( D×t×(μ_i-1)/d ) ]

no. of fringes crossing a point on the screen =   [ (  D×t×(μ_f-1)/d )  -  ( D×t×(μ_i-1)/d ) ] / [ λ×D/d ] = t×μ×α×ΔT/λ  ...................(4)

where α is temperature coefficient of refractive index and ΔT is change in temperature.

 by substituting the values t = 2×10^-2 m , ΔT = 5 °C and the respective values for μ and λ in eqn.(4), we get

no. of fringes crossing a point on the screen is  1/2

Qn.(9)

 Intensity I = I₁ + I₂ + 2 (I₁ I₂ )^1/2cosδ

 If I₁ = I and I  = I   , then by using cosδ=+1, we have maximum intensity I₀ = 4I

If I₁ = I and I  = I/2  , then by using cosδ=+1, we have maximum intensity I' = I   [ (3/2) + (2/√2) ] = (I₀ / 4)[ (3/2) + √2 ]