In YDSE- let A and B be two slits- Films of thicknesses tA and tB and refractive indices -A and -B are placed in front of A and B- respectively- If -A tA - -BtB- then the central maxima will -

The correct option is D (b) if t_B > t_A and (c) if t_B Path difference introduced in ray from slit A due to the film =x_1=t_A(μ_A 1) Path diffe ...

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Byju's Answer

Standard XII

Physics

Optical Path Length

In YDSE, let ...

Question

In YDSE, let $A$ and $B$ be two slits. Films of thicknesses $t_{A}$ and $t_{B}$ and refractive indices $μ_{A}$ and $μ_{B}$ are placed in front of $A$ and $B$ , respectively. If $μ_{A} t_{A} = μ_{B} t_{B}$ , then the central maxima will :

Not shift

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Shift toward

A

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Shift toward

B

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(b) if

t_{B} > t_{A}

and (c) if

t_{B} < t_{A}

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Solution

The correct option is D (b) if $t_{B} > t_{A}$ and (c) if $t_{B} < t_{A}$
Path difference introduced in ray from slit A due to the film $= x_{1} = t_{A} (μ_{A} - 1)$
Path difference introduced in ray from slit B due to the film $= x_{2} = t_{B} (μ_{B} - 1)$
If $t_{B} > t_{A}$ ,
Path difference between the rays= $x_{1} - x_{2} = t_{B} - t_{A} > 0$ (Since $t_{A} μ_{A} = t_{B} μ_{B}$ )
=path difference at a point 'y' on the screen to due extra path traveled by the light from source B $= \frac{y d}{D} > 0$
$⟹ y > 0$
Therefore the shift is towards A.
If $t_{B} < t_{A}$
$x_{1} - x_{2} = t_{B} - t_{A} < 0$
Thus $y < 0$
Thus the shift is towards B.

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In YDSE, let A and B be two slits. Films of thicknesses tA and tB and refractive indices μA and μB are placed in front of A and B, respectively. If μAtA=μBtB, then the central maxima will :

In YDSE, let $A$ and $B$ be two slits. Films of thicknesses $t_{A}$ and $t_{B}$ and refractive indices $μ_{A}$ and $μ_{B}$ are placed in front of $A$ and $B$ , respectively. If $μ_{A} t_{A} = μ_{B} t_{B}$ , then the central maxima will :