A disc is placed on the surface of pond filled with liquid of refractive index `(5)/(3)`. A source of light is placed 4m below the surface of liquid. Calculate the minimum area of the disc so that light does not come out of liquid.

Explanation

`sinC=1/mu`
or, `r/(sqrt(r^(2)+h^(2)))=1/mu`
or, `mu=(sqrt(r^(2)+h^(2)))/r`
or, `mu^(2)=(r^(2)+h^(2))/r^(2)=1+h^(2)/r^(2)`
or, `h^(2)/r^(2)=mu^(2)-1`
or, `r=h/(sqrt(mu^(2)-1))=4/(sqrt(25/9-1))=4/(4/3)=3m`
`therefore` Required area of the disc = `pir^(2)=3.14xx9m^(2)=28.28m^(2)`