Question

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

Answer 1

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