In YDSE the separation between slits S1 and S2 is 5mm and slits are illuminating with light of wavelength 5000A0 . The interference fringes are formed on the screen with central maximum at ‘O’. A glass plate of μ=32 with thickness 1μm at end A and is linearly increasing to 7μm at end B is moved in front of slits with constant speed 1mm/sec downwards. At t=0, end A is right in front of S1 . AB=6mm

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In YDSE the separation between slits $S_{1}$ and $S_{2}$ is 5mm and slits are illuminating with light of wavelength $5000 A^{0}$ . The interference fringes are formed on the screen with central maximum at ‘O’. A glass plate of $μ = \frac{3}{2}$ with thickness $1 μ m$ at end A and is linearly increasing to $7 μ m$ at end B is moved in front of slits with constant speed $1 m m / \sec$ downwards. At t=0, end A is right in front of $S_{1}$ . AB=6mm

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Order of maxima at O when B is in front of slit 1 is 7.

Order of maxima at O when B is in front of slit 1 is 5.

At t=0, just after end A of plate comes in front of slit 1, the order of maxima at O is 1.

At $t = 5$ sec $5^{th}$ bright fringe is at O

answer is A, B, C.

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$y = α x + C 1 μ m = α \times 0 + C \Rightarrow C = 1 μ m 7 μ m = α \times 6 m m + 1 μ m$

$\Rightarrow m = 10^{- 3} S_{1} S_{2} = d = 5 m m, λ = 5000 \overset{°}{A} μ = 1.5$

(A)   $\frac{(μ - 1) t}{λ} = \frac{0.5 \times 1 \times 10^{- 6}}{5000 \times 10^{- 10}} = 1$
(B) At t = 5s,    $x = \frac{1 m m}{s} \times 5 s = 5 m m$
   $y = 5 μ m + 1 μ m = 6 μ m$
No. of fringes crossed    $\frac{0.5 \times 6 \times 10^{- 6}}{5000 \times 10^{- 10}} = 6$
6^th bright fringe will be at O
(C,D) Here $x = 6 m m$
$\Rightarrow y = 6 μ m + 1 μ m = 7 μ m$
No. of fringes crossed = $\frac{0.5 \times 7 \times 10^{- 6}}{5000 \times 10^{- 10}} = 7$