Question

# A disc is placed on the surface of a pond which has a refractive index $\frac{5}{3}$. A source of light is placed $4m$ below the surface of a liquid. Find the minimum radius of a disc so that light does not come out from the disc?

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Solution

## Step 1: Given dataThe depth is $4m$and refractive index $\mu =\frac{5}{3}$Step 2: Formula used$\mathrm{sin}a=\frac{1}{\mu }\left[\mu =refractiveindex\right]$Step 3: Calculating the minimum radiusFirstly we will draw the diagram and then we will proceed with the solution. Here the depth is given from where the luminous object or source light is kept. And a disc is placed so that the light does not come outLet us assume the radius of the disc is $PR=x$ .$\mathrm{sin}a=\frac{1}{\mu }=\frac{1}{\frac{5}{3}}=\frac{3}{5}$And from the figure, we can say that, $\mathrm{sin}a$ is nothing but perpendicular upon hypotenuse.$\frac{PR}{OR}=\frac{3}{5}$On comparing,$PR=x=3m$Hence, the minimum radius of a disc so that light does not come out from a disc is 3m$\phantom{\rule{thinmathspace}{0ex}}$.

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