In a regular YDSE, when thin film of refractive index μ is placed in front of the lower slit then it is observed that the intensity at the central point becomes half of the original intensity. It is also observed that the initial 3rd maxima is now below the central point and the initial 4th minima is above the central point. Now, a film of refractive index μ1 and thickness same as the above film, is put in the front of the lower slit also. It is observed that whole fringe pattern shifts by one fringe width. What is the value of μ1?

Engineering

Physics

Wave Optics

Question

In a regular YDSE, when thin film of refractive index μ is placed in front of the lower slit then it is observed that the intensity at the central point becomes half of the original intensity. It is also observed that the initial 3^rd maxima is now below the central point and the initial 4^th minima is above the central point. Now, a film of refractive index μ₁ and thickness same as the above film, is put in the front of the lower slit also. It is observed that whole fringe pattern shifts by one fringe width. What is the value of μ₁?

(4μ+9)/12

(4μ+9)/11

(4μ+9)/13

None

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Δφ = $\frac{2 π}{λ}$ t (μ – 1)   ... (1)
6π < Δφ < 7π        ... (2)
   I = I₀ $\cos^{2} (\frac{Δφ}{2})$
   I = I₀ / 2
⟹ $\cos (\frac{Δφ}{2})$ = ± $\frac{1}{\sqrt{2}}$ ⟹ $\frac{Δφ}{2}$ = 2nπ ± $\frac{π}{4}$
Δφ = 4nπ ± π / 2   ... (3)
Solving equation (2) & (3) we get
Δφ = 6.5 π
from (1)
   6.5 π = $\frac{2 π}{λ}$ t (μ – 1)   ... (4)
Introducing record film shifted the fringe pattern by one fringe with means.
       t (μ₁ – 1) = λ     ... (5)
Solving 4 & 5
⟹ $μ_{1} = \frac{4 μ + 9}{13}$

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