An observer can see through a pin-hole the top end of a thin rod of height $h$ , placed as shown in the figure. The beaker height is $3 h$ and its radius is $h$ . When the beaker is filled with a liquid up to a height $2 h$ , he can see the lower end of the rod. Then the refractive index of the liquid is [2002]

Question

Smart Achievers · Accepted Answer

(2)

For the image of point P of the rod PQ to be seen by the observer, it should be formed at point Q.

In $h$ ,

$N S = Q S = 2 h$

$∴$ $∠ N Q S = 45 °$

$∴$ $r = 45 °$

Now in $∆ Q M A$ ,

$∠ M Q A = 45 °$

$∴$ $M A = Q A = h$

${}_{2}^{1}μ = \frac{\sin r}{\sin i} = \frac{\sin 45 °}{\sin i}$ $...(i)$

In $∆ P M B$ ,

$P M^{2} = 4 h^{2} + h^{2} = 5 h^{2}$

$∴$ $\sin i = \frac{h}{\sqrt{5} h} = \frac{1}{\sqrt{5}}$ $...(ii)$

From eq. (i) and (ii) $\Rightarrow {}_{2}^{1}μ = \frac{\sin 45 °}{\sin i} = \sqrt{\frac{5}{2}}$

Solution

1 $\frac{5}{2}$

2 $\sqrt{\frac{5}{2}}$

3 $\sqrt{\frac{3}{2}}$

4 $\frac{3}{2}$

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Solution

1 52

2 52

3 32

4 32

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1 $\frac{5}{2}$

2 $\sqrt{\frac{5}{2}}$

3 $\sqrt{\frac{3}{2}}$

4 $\frac{3}{2}$