Question 266: Waves Optics

Question

In Young's double slit experiment, the distance between screen and aperture is 1 m . The slit width is 2 mm . Light of $6000 \mathop {\rm{A}}\limits^{\rm{o}}$ is used. If a thin glass plate ( $\mu=1.5$ ) of thickness 0.04 mm is placed over one of the slits, then there will be a lateral displacement of the fringes by

Accepted Answer

We are given a glass plate with thickness $t=0.04~\mathrm{mm}$.
When this glass plate is placed on one slit, the fringe pattern shifts. The amount the pattern shifts (lateral displacement) is found using:
$$ d' = \frac{t(\mu - 1)D}{d} $$
Here's what the variables mean:
$t$ = Thickness of glass plate ($0.04~\mathrm{mm}$)
$\mu$ = Refractive index of glass ($1.5$)
$D$ = Distance from slits to screen ($1~\mathrm{m}$)
$d$ = Distance between the slits ($2~\mathrm{mm}$)
Let's put in the numbers. First, change all measurements to meters:
$t = 0.04~\mathrm{mm} = 0.04 	imes 10^{-3}~\mathrm{m}$
$d = 2~\mathrm{mm} = 2 	imes 10^{-3}~\mathrm{m}$
Now substitute the values into the formula:
$$ d' = \frac{0.04 	imes 10^{-3} 	imes (1.5 - 1) 	imes 1}{2 	imes 10^{-3}} $$
Calculate $1.5 - 1 = 0.5$.
So,

$$ d' = \frac{0.04 	imes 10^{-3} 	imes 0.5 	imes 1}{2 	imes 10^{-3}} $$
Multiply the top: $0.04 	imes 0.5 = 0.02$. So the top becomes $0.02 	imes 10^{-3}$.
Now divide:

$$ d' = \frac{0.02 	imes 10^{-3}}{2 	imes 10^{-3}} $$
Divide $0.02$ by $2$ to get $0.01$.
So,

$$ d' = 0.01~\mathrm{m} $$
$0.01~\mathrm{m} = 1~\mathrm{cm}$
Therefore, the fringe pattern shifts by $1~\mathrm{cm}$.

Answer

0.5 cm

Answer

1 cm

Answer

1.5 cm

Answer

2 cm

Waves Optics

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