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Standard X

Physics

Lateral Displacement

A stationary ...

Question

A stationary swimmer $S$ , inside a liquid of refractive index, $μ_{1}$ , is at distance $d$ from a fixed point $P$ inside the liquid. A rectangular block of width $t$ and refractive index $μ_{2} (μ_{2} < μ_{1})$ is now placed between $S$ and $P$ . $S$ will observe $P$ to be at a distance

d - t (\frac{{2 μ}_{1}}{μ_{2}} - 1)

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d - t (1 - \frac{μ_{2}}{{3 μ}_{1}})

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d - t (1 - \frac{μ_{2}}{μ_{1}})

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d - t (\frac{μ_{1}}{μ_{2}} - 1)

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Solution

The correct option is C $d - t (1 - \frac{μ_{2}}{μ_{1}})$
Refractive index of block with respect to medium $= \frac{μ_{2}}{μ_{1}}$ when the block is introduced, $P$ will appear to move towards $S$ through a distance $t^{'} = t (1 - \frac{μ_{2}}{μ_{1}})$ ; here $t^{'} < 0$ .
$∴$ apparent distance from $S$ to $O^{'} = d - t^{'} = d - t (1 - \frac{μ_{2}}{μ_{1}})$

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A stationary swimmer S, inside a liquid of refractive index, μ1, is at distance d from a fixed point P inside the liquid. A rectangular block of width t and refractive index μ2(μ2<μ1) is now placed between S and P. S will observe P to be at a distance

The correct option is C d−t(1−μ2μ1)Refractive index of block with respect to medium=μ2μ1 when the block is introduced, P will appear to move towards S through a distance t′=t(1−μ2μ1); here t′<0.∴ apparent distance from S to O′=d−t′=d−t(1−μ2μ1)

A stationary swimmer $S$ , inside a liquid of refractive index, $μ_{1}$ , is at distance $d$ from a fixed point $P$ inside the liquid. A rectangular block of width $t$ and refractive index $μ_{2} (μ_{2} < μ_{1})$ is now placed between $S$ and $P$ . $S$ will observe $P$ to be at a distance