Question

A convex lens is held $45cm$ above the bottom of an empty tank. The image of a point on the bottom of a tank is formed $36cm$ above the lens. Now liquid is poured into the tank to a depth of $40cm$. It is found that the distance of the image of the same point on the bottom of the tank is $40cm$ above the lens. The refractive index of the liquid is $\frac{8}{a}$ then value of $a$ is.

Answer 1

Answer: 7.00

Initially from lens
$u=-45cm,v=+36cm$
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{36}-\frac{1}{-45}=\frac{1}{f}$
$\frac{1}{f}=\frac{5+4}{180}$
$\Rightarrow f=20cm$
When liquid is filled, for water surface
$u=-40cm,R=\infty ,{\mu }_1={\mu }_{,}{\mu }_2=1$
$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$
$\frac{1}{v}-\frac{\mu }{-40}=\frac{1-\mu }{\infty }$
$\Rightarrow v=\frac{-40}{\mu }<0$
so for lens, $u=-\left[5+\frac{40}{\mu }\right]=-\left[\frac{5\mu +40}{\mu }\right]$
$v=48cm,f=20cm$
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{48}-\frac{\mu }{-[5\mu +40]}=\frac{1}{20}$
$\frac{\mu }{5\mu +40}=\frac{1}{20}-\frac{1}{48}=\frac{12-5}{240}$
$240\mu =35\mu +280$
$205\mu =280$
$\Rightarrow \mu =\frac{280}{205}=1.3658$