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In Young's double slit experiment, the fringe width is found to be $0.4 \mathrm{~mm}$. If the whole apparatus is immersed in a liquid of refractive index $\frac{4}{3}$ without changing geometrical arrangement, the new fringe width will be
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0.30 mm
If fringe widths of air and water are $\beta_{\text {air }}$ and $\beta_{\text {water }}$ respectively, then fringe width, $\beta_{\text {water }}=\frac{\beta_{\text {air }}}{\mu}$
Given, $\mu=\frac{4}{3}, \beta_{\text {air }}=0.4$
Hence, $\beta_{\text {water }}=\frac{0.4}{4 / 3}=0.3 \mathrm{~mm}$
Given, $\mu=\frac{4}{3}, \beta_{\text {air }}=0.4$
Hence, $\beta_{\text {water }}=\frac{0.4}{4 / 3}=0.3 \mathrm{~mm}$
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