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How does the fringe of interference frin...

How does the fringe of interference fringes change, when the whole apparatus Young's experiment is kept in water (refract index 4/3)?

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Fringe width, `beta = (D lamda)/(d) rArr beta = lamda` for same D and d. When the whole aparatus is immersed in a transparent liquid of refractive index n = 4/3, the wavelength decreases to `lamda' = lamda/n = (lamda)/(4//3)` So, fringe width decreases to `3/4` times.
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