Question

Figure $9.37$ shows an equiconvex lens [of reflective index $1.50$] in contact with a liquid layer on top of a plane mirror. A small needle with its up on the principle axis is moved along the axis unit its inverted image is found at the position of the needle.The distance of the needle from the lens is measured to be $45.0cm$ The liquid is removed and the experiments is repeated. The new distance is measured to be $30.0cm$. What is the reflexive index of the liquid?

Answer 1

Let focal length of convex lens (glass) ,$f_1=30cm$

focal length of plane concave lens (liquid) is$f_2$

and it is given that combined focal length , $F=45cm$

we know,

$\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{F}$ [ combined focal length formula]

so, $\frac{1}{f_2}=\frac{1}{F}-\frac{1}{f_1}$

$\frac{1}{f_2}=\frac{1}{45}-\frac{1}{30}$

$f_2=\frac{(2-3)}{90}=-\frac{1}{90}$

$f_2=-90cm$

for lens (glass),

$R_1=R$and $R_2=-R$

use lens maker formula,

$\frac{1}{f_1}=(u-1)[\frac{1}{R_1}-\frac{1}{R_2}]$

$\frac{1}{30}=(\frac{3}{2}-1)[\frac{1}{R}+\frac{1}{R}]$

$\frac{1}{30}=\frac{1}{2\times 2/R}=\frac{1}{R}$

$R=30cm$

now, for plane concave lens (liquid)

$R_1=-R=-30cm$ , $R_2=\infty$

so, $\frac{1}{f_2}=(u^{\prime }-1)[\frac{1}{R1}-\frac{1}{R2}]$

$\frac{1}{-90}=(u^{\prime }-1)[\frac{1}{-30}-\frac{1}{\infty }]$

$-\frac{1}{90}=\frac{(u^{\prime }-1)}{-30}$

$\frac{1}{3}=u^{\prime }-1$

$u^{\prime }=\frac{4}{3}$

hence, refractive index of liquid is $\frac{4}{3}$