(B) The optical path difference introduced by the slab is $\Delta x = (\mu - 1)t$.Given $\mu = 1.5$ and $t = d/4$,the path difference is $\Delta x = (

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(B) The optical path difference introduced by the slab is $\Delta x = (\mu - 1)t$.Given $\mu = 1.5$ and $t = d/4$,the path difference is $\Delta x = (1.5 - 1) \times (d/4) = 0.5 \times (d/4) = d/8$.For the central maxima to form at a position $y$ on the screen,the path difference at that point must be zero.The path difference at point $y$ is $\Delta x_{net} = \frac{yd}{D} - \Delta x = 0$.Therefore,$\frac{yd}{D} = \frac{d}{8}$.Solving for $y$,we get $y = D/8 = 0.125\ D$.Since the slab is placed in the path $AS_2$,the light through $S_2$ is delayed,causing the central maxima to shift towards the side of the slab,which is below $O$.

Class 12 Physics Wave Optics Variations in YDSE (Young's Double Slit Experiment)

Difficult

In a standard $YDSE$ setup,a small transparent slab of thickness $t$ and refractive index $\mu = 1.5$ is placed along the path $AS_2$ (as shown in the figure). Given that the slab thickness $t = d/4$,where $d$ is the slit separation,and the distance from the source $A$ to the slits is not explicitly needed for the shift calculation,find the position of the central maxima on the screen relative to $O$. Assume the distance between the slits and the screen is $D$.

A
$0.125\ D\ m$ above $O$
B
$0.125\ D\ m$ below $O$
C
$0.25\ D\ m$ below $O$
D
$0.25\ D\ m$ above $O$

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Wave Optics

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Variations in YDSE (Young's Double Slit Experiment)

In a Young's double-slit experiment,let $A$ and $B$ be the two slits. Thin films of thicknesses $t_A$ and $t_B$ and refractive indices $\mu_A$ and $\mu_B$ are placed in front of slits $A$ and $B$ respectively. If $\mu_A t_A = \mu_B t_B$,the central maximum will:

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$A$ parallel beam of light of wavelength $500 \ nm$ is incident at an angle $30^o$ with the normal to the slit plane in a Young's double-slit experiment. The intensity due to each slit is $I_o$. Point $O$ is equidistant from $S_1$ and $S_2$. The distance between the slits is $1 \ mm$.

Difficult

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To make the central fringe appear at the centre $O,$ a mica sheet of refractive index $\mu = 1.5$ is introduced. Choose the correct statement:

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The central fringe of the interference pattern produced by light of wavelength $6000 \mathring A$ is found to shift to the position of the $4^{\text {th}}$ bright fringe after a glass plate of refractive index $1.5$ is introduced in front of one slit in Young's experiment. The thickness of the glass plate will be ......... $\mu m$.

Medium

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In the figure shown in $YDSE$, a parallel beam of light is incident on the slits from a medium of refractive index $n_1$. The wavelength of light in this medium is $\lambda_1$. $A$ transparent slab of thickness $t$ and refractive index $n_3$ is placed in front of one slit. The medium between the screen and the plane of the slits is $n_2$. The phase difference between the light waves reaching point $O$ (symmetrical, relative to the slits) is:

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In a Young's double-slit experiment,let $A$ and $B$ be the two slits. Thin films of thicknesses $t_A$ and $t_B$ and refractive indices $\mu_A$ and $\mu_B$ are placed in front of slits $A$ and $B$ respectively. If $\mu_A t_A = \mu_B t_B$,the central maximum will:

$A$ parallel beam of light of wavelength $500 \ nm$ is incident at an angle $30^o$ with the normal to the slit plane in a Young's double-slit experiment. The intensity due to each slit is $I_o$. Point $O$ is equidistant from $S_1$ and $S_2$. The distance between the slits is $1 \ mm$.

To make the central fringe appear at the centre $O,$ a mica sheet of refractive index $\mu = 1.5$ is introduced. Choose the correct statement:

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