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Physics · 2025 · 2 marks
CBSE 2025 · Region 2 · Set 1 · Q18
In a double-slit experiment, 6^th dark fringe is observed at a certain point of the screen. A transparent sheet of thickness t and refractive index n…
In a double-slit experiment, $\displaystyle 6^{\text {th }}$ dark fringe is observed at a certain point of the screen. A transparent sheet of thickness t and refractive index n is now introduced in the path of one of the two interfering waves to increase its phase by $\displaystyle 2 \pi(\mathrm{n}-1) \mathrm{t} / \lambda$. The pattern is shifted and $\displaystyle 8^{\text {th }}$ bright fringe is observed at the same point. Find the relation for thickness t in terms of n and $\displaystyle \lambda$.
Marking-scheme solution
Phase difference for $\displaystyle 6^{\text {th }}$ dark fringe $\displaystyle =11 \pi$
Phase difference for $\displaystyle 8^{\text {th }}$ bright fringe $\displaystyle =16 \pi$
$\displaystyle \Delta \phi+\phi_{6}=\phi_{8}$
$\displaystyle \frac{2 \pi(n-1) t}{\lambda}+11 \pi=16 \pi$
$\displaystyle t=\frac{5 \lambda}{2(n-1)}$
Path difference for $\displaystyle 6^{\text {th }}$ dark fringe $\displaystyle \left(x_{6}\right)=\frac{11 \lambda}{2}$
Path difference for $\displaystyle 8^{\text {th }}$ bright fringe $\displaystyle \left(x_{8}\right)=8 \lambda$
Increase in path difference after introducing the sheet $\displaystyle (x)=\frac{\lambda}{2 \pi} \times \frac{2 \pi(n-1) t}{\lambda}=(n-1) t$
$\displaystyle \therefore x_{6}+x=x_{8}$
$\displaystyle \frac{11 \lambda}{2}+(n-1) t=8 \lambda$
On solving:
$\displaystyle t=\frac{5 \lambda}{2(n-1)}$
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