AP EAMCET · PHYSICS · Ray Optics
The lower half of a vessel of depth \(2 d \mathrm{~cm}\) is filled with a liquid of refractive index \(\mu_1\) and the upper half with a liquid of refractive index \(\mu_2\). The apparent depth of the vessel seen perpendicularly is
- A \(d\left(\frac{\mu_1 \mu_2}{\mu_1+\mu_2}\right)\)
- B \(d\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}\right)\)
- C \(2 d\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}\right)\)
- D \(2 d\left(\frac{1}{\mu_1 \mu_2}\right)\)
Answer & Solution
Correct Answer
(B) \(d\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}\right)\)
Step-by-step Solution
Detailed explanation
If immiscible liquids of refractive indices \(\mu_1\) and \(\mu_2\) are filled in a vessel and their real depths are \(d_1\) and \(d_2\) as shown below Then, apparent depth of the vessel seen perpendicularly is given as \(d_{\mathrm{app}}=\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}\)…
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