Question 74: Waves Optics

Question

In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness $t$ and refractive index $n(=1.5)$, the central fringe shifts by 0.2 cm . The value of $t$ is
$\_\_\_\_$ cm.

Accepted Answer

In a Young's Double Slit Experiment (YDSE), the position of the fringes is determined by the path difference between the two waves. When a transparent sheet of thickness t and refractive index n is placed in front of one slit, the light travels through the medium instead of air for that distance.
The optical path through the sheet is nt , while the path in air for the same distance would be t . Thus, an additional path difference $(\Delta \mathrm{x})$ is introduced :
$$ \Delta \mathrm{x}_{	ext {optical }}=\mathrm{nt}-\mathrm{t}=(\mathrm{n}-1) \mathrm{t} $$
The central fringe which originally occurs at the centre where the geometric path difference is zero, shifts to a new position $\Delta y$ where the net path difference is zero.
The geometric path difference for a point at distance $y$ from the center is :
$$ \Delta \mathrm{x}_{\mathrm{geo}}=\frac{\mathrm{d} \cdot \mathrm{y}}{\mathrm{D}} $$
Where d is the distance between slits and D is the distance to the screen.
So, at the position of central maximum net path difference must be zero, $\Delta \mathrm{x}_{	ext {optical }}+\Delta \mathrm{x}_{	ext {geo }}=0$
$$ \frac{\mathrm{d} \cdot \mathrm{y}}{\mathrm{D}}=(\mathrm{n}-1) \mathrm{t} $$
$$ \Rightarrow \mathrm{t}=\frac{\mathrm{y} \cdot \mathrm{~d}}{\mathrm{D}(\mathrm{n}-1)} $$
Putting the values,
Distance between slits, $\mathrm{d}=0.1 \mathrm{~cm}$
Distance to screen $\mathrm{D}=50 \mathrm{~cm}$
Shift of central fringe $y=0.2 \mathrm{~cm}$
Refractive index $\mathrm{n}=1.5$
$$ \mathrm{t}=\frac{0.2 	imes 0.1}{50(1.5-1)} $$
$$\Rightarrow $$ $t=\frac{0.02}{25}$
$$\Rightarrow $$ $$ \mathrm{t}=8 	imes 10^{-4} \mathrm{~cm} $$
Therefore, the thickness of the sheet to be $8 	imes 10^{-4} \mathrm{~cm}$.
Hence, the correct option is (B).

Answer

$6.0 	imes 10^{-3}$

Answer

$8 	imes 10^{-4}$

Answer

$5.0 	imes 10^{-3}$

Answer

$5.6 	imes 10^{-4}$

Waves Optics

In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness t and refractive index n(=1.5), the central fringe shifts by 0.2 cm . The value of t is \_\_\_\_ cm.

Choose the correct answer: