KCET · Physics · Ray Optics
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
- 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 \(\frac{n d}{d+\sqrt{2 n}}\)
Answer & Solution
Correct Answer
(B) \(\frac{d(n+\sqrt{2})}{n \sqrt{2}}\)
Step-by-step Solution
Detailed explanation
Refractive index
\[
\mu=\frac{\text { Real depth }(d)}{\text { Apparent depth }(x x)}
\]
For 1st liquid, \(\sqrt{2}=\frac{d}{x_{1}}\)
\(\Rightarrow \quad x_{1}=\frac{d}{\sqrt{2}}\)
Similarly, for 2 nd liquid,
\[
n=\frac{d}{x_{2}} \Rightarrow x_{2}=\frac{d}{n}
\]
Total apparent depth \(=x_{1}+x_{2}\)
\[
=\frac{d}{\sqrt{2}}+\frac{d}{n}=\frac{d(n+\sqrt{2})}{n \sqrt{2}}
\]
\[
\mu=\frac{\text { Real depth }(d)}{\text { Apparent depth }(x x)}
\]
For 1st liquid, \(\sqrt{2}=\frac{d}{x_{1}}\)
\(\Rightarrow \quad x_{1}=\frac{d}{\sqrt{2}}\)
Similarly, for 2 nd liquid,
\[
n=\frac{d}{x_{2}} \Rightarrow x_{2}=\frac{d}{n}
\]
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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