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Answer: 3. 1.78

Key idea: compare fringe positions using bright fringe \(y_n=\dfrac{n\lambda D}{d}\) and dark fringe \(y_n=\dfrac{(2n-1)\lambda D}{2d}\).

In air, 5th dark fringe: \(y=\dfrac{9}{2}\dfrac{\lambda D}{d}=4.5\dfrac{\lambda D}{d}\).

In the medium, wavelength is \(\lambda/\mu\). 8th bright fringe: \(y=8\dfrac{\lambda D}{\mu d}\).

Equating: \(\dfrac{8}{\mu}=4.5\Rightarrow \mu=\dfrac{8}{4.5}=1.78\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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