A vessel of height 2 d is half filled with a liquid of refractive...

A vessel of height

2 d

is half filled with a liquid of refractive index

\sqrt{2}

and the other half with a liquid of refractive index

n

(the given liquids are immiscible). Then, the apparent depth of the inner surface of the bottom of the vessel (neglecting the thickness of the bottom of the vessel) will be

Options

A $\frac{n}{d(n+\sqrt{2})}$
B $\frac{d(n+\sqrt{2})}{n \sqrt{2}}$
C $\frac{\sqrt{2} n}{d(n+\sqrt{2})}$
D $\frac{n d}{d+\sqrt{2 n}}$

Correct answer

B. $\frac{d(n+\sqrt{2})}{n \sqrt{2}}$

Step-by-step solution

Refractive index

\mu=\frac{\text{ Real depth }(d)}{\text{ Apparent depth }(x)}

For 1st liquid,

\sqrt{2}=\frac{d}{x_1} \Rightarrow \quad x_1=\frac{d}{\sqrt{2}}

Similarly, for 2nd liquid,

n=\frac{d}{x_2} \Rightarrow \quad x_2=\frac{d}{n}

Total apparent depth

=x_1+x_2 =\frac{d}{\sqrt{2}}+\frac{d}{n}=\frac{d(n+\sqrt{2})}{n \sqrt{2}}

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