Three slabs $A, B$ and $C$ are placed in front of slits in $Y D S E$ as shown in figure. Refractive index of slabs $A, B$ and $C$ are $4 / 3, 5 / 3$ and $8 / 3$ respectively and thickness of $A, B$ and $C$ are $2 m m, 1 m m$ and $1.5 m m$ respectively. A liquid of refractive index $2 μ / 3$ is filled between slits and screen . A monochromatic light of wavelength $5000 ˚ A$ is incident on the slits. If the central bright frings is obtained at $C$ then find the value of $^{'} μ^{'}$ .

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Solution

The correct option is A $5$
$S_{1} C / o p t = \frac{4}{3} S_{1} C + μ_{B} \times t_{B} - t_{B} = S_{1} C - (μ_{B} - 1) t_{B}$
$S_{2} C / o p t = \frac{4}{3} (S_{2} C - t_{C} + μ_{C} t_{C} = S_{2} C + (μ_{C} - 1) \times t_{C}$
$Δ x = 0$ $\Rightarrow$ $(μ_{A} t_{A} - t_{A}) + S_{1} C + (μ_{B} - 1) t_{B} - S_{2} C - (μ_{C} - 1) \times t_{C}$
$(μ_{A} - 1) t_{A} + (μ_{A} - 1) t_{B} = ((μ_{C} - 1) \times t_{C}$
$(\frac{4}{3} - 1) \times 2 + (\frac{5}{3} - 1) \times 1 = (\frac{8}{3} - 1) \times 15$
optical path of above ray $= μ_{A} t_{A} + μ_{B} t_{B} + (S_{1} C - t_{B}) \times \frac{2 μ}{3}$
optical path of ray below $= t_{A} + μ_{C} t_{C} + (S_{2} C - t_{C}) \times \frac{2 u}{3}$
$∴$ As path difference is zero
$μ_{A} t_{A} + μ_{B} t_{B} + (S_{1} C - t_{A}) \times \frac{2 μ}{3} = t_{A} + μ_{C} t_{C} + (S_{2} C - t_{C}) \times \frac{2 μ}{3}$
$\frac{4}{3} \times 2 + \frac{5}{3} \times 1 + S_{1} C \times \frac{2 μ}{3} - \frac{2 μ}{3} \times t_{B} = 2 + \frac{8}{3} \times 1.5 + S_{2} C \times \frac{2 μ}{3} - 1.5 \times \frac{2 μ}{3}$
$\frac{8}{3} + \frac{5}{3} - \frac{2 μ}{3} \times 1 = 2 + 4 - μ$
$\frac{13}{3} - \frac{2 μ}{3} = 6 - μ = \frac{13}{3} - 6 = - (\frac{2 μ}{3} - μ) = - \frac{μ}{3}$
$\frac{13 - 18}{3} = - \frac{μ}{3} = - \frac{5}{3} = - \frac{μ}{3}$
$μ = 5$

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Q. In a YDSE setup light of wavelength

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2 m

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(thickness

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(A) If slab $1$ is placed in front of slit $S_{1}$	(P) Central maxima at $C$
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(C) If slab $3$ is placed in front of slab $1$ along with condition (B)	(R) Fringe width is equal to $0.8 m m$
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