A liquid of refractive index n is filled in a tank. A plane mirror is kept at the bottom of the tank. A point like object (P) is kept at a height h from the mirror on the liquid surface. As observer observes the object and its image in the veryically downward direction from top. How much distance will observer note between P and its image?

2 n - h

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\frac{2 h}{n}

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\frac{2 h}{(n - 1)}

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(1 + \frac{1}{n})

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Solution

The correct option is B $\frac{2 h}{n}$
$Shift = \frac{(H - h)}{μ} = K_{1}$
$\begin{matrix} Similarly for plane mirror K_{2} = \frac{(H + h)}{μ} K(apparent) = K_{2} - K_{1} = \frac{2 h}{μ} = \frac{2 h}{n} \end{matrix}$

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The correct option is B 2hn Shift =(H−h)μ=K1 Similarly for plane mirror K2=(H+h)μ K(apparent) =K2−K1=2hμ=2hn

The correct option is B $\frac{2 h}{n}$
$Shift = \frac{(H - h)}{μ} = K_{1}$
$\begin{matrix} Similarly for plane mirror K_{2} = \frac{(H + h)}{μ} K(apparent) = K_{2} - K_{1} = \frac{2 h}{μ} = \frac{2 h}{n} \end{matrix}$