An equiconvex lens with radii of curvature of magnitude r each, is put over a liquid layer poured on top of a plane mirror. A small needle, with its tip on the principal axis of the lens, is moved along the axis until its inverted real image coincides with the needle itself. The distance of the needle from the lens is measured to be 'a'. On removing the liquid layer and repeating the experiment the distance is found to be 'b'.

Given that two values of distances measured represent the focal length values in the two cases, obtain a formula for the refractive index of the liquid.

If r = 10cm, a = 15cm, b = 10cm, find the refractive index of the liquid.

Q: 1. An equiconvex lens with radii of curvature of magnitude r each, is put over a liquid layer poured on top of a plane mirror. A small needle, with its tip on the principal axis of the lens, is moved along the axis until its inverted real image coincides with the needle itself. The distance of the needle from the lens is measured to be 'a'. On removing the liquid layer and repeating the experiment the distance is found to be 'b'. Given that two values of distances measured represent the focal length values in the two cases, obtain a formula for the refractive index of the liquid. 2. If r = 10cm, a = 15cm, b = 10cm, find the refractive index of the liquid.

1. The focal length $(f_1)$ of lens is given by $\frac{1}{\text{f}_1}=(\text{n}-1)(\frac{1}{\text{r}}+\frac{1}{\text{r}})=\frac{2(\text{n}-1)}{\text{r}}$ Given $f_1= b$ $\Rightarrow\frac{1}{\text{b}}=\frac{2(\text{n}-1)}{\text{r}}$ $\Rightarrow\text{b}=\frac{\text{r}}{2(\text{n}-1)}$ The focal lenght of liquid lens (plano concave lens) is $\frac{1}{\text{f}_2}=(\text{n}_{\text{l}}-1)(-\frac{1}{\text{r}}-\frac{1}{\propto})$ $=-\frac{(\text{n}_{\text{l}}-1)}{\text{r}}$ $\Rightarrow\text{f}_2=\frac{\text{r}}{(\text{n}_{\text{l}}-1)}$ As glass lens and liquid lens are in contact $\therefore\frac{1}{\text{f}}=\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}=\frac{1}{\text{b}}-\frac{(\text{n}_{\text{l}}-1)}{\text{r}}$ Given $f = a$ $\therefore\frac{1}{\text{a}}=\frac{1}{\text{b}}-\frac{\text{n}_{\text{l}}-1}{\text{r}}$ $\Rightarrow\frac{\text{n}_{\text{l}}-1}{\text{r}}=\frac{1}{\text{b}}-\frac{1}{\text{a}}$ $\Rightarrow\text{n}_{\text{l}}-1=\text{r}\big(\frac{1}{\text{b}}-\frac{1}{\text{a}}\big)$ Refractive index of liquid, 2. $\text{n}_{\text{l}}=1+\frac{\text{r}}{\text{b}}-\frac{\text{r}}{\text{a}}$ $\text{n}_{\text{l}}=1+\frac{10}{10}-\frac{10}{15}$ $=1+1-\frac{2}{3}=2-\frac{2}{3}=\frac{4}{3}=1.33$

Question

An equiconvex lens with radii of curvature of magnitude r each, is put over a liquid layer poured on top of a plane mirror. A small needle, with its tip on the principal axis of the lens, is moved along the axis until its inverted real image coincides with the needle itself. The distance of the needle from the lens is measured to be 'a'. On removing the liquid layer and repeating the experiment the distance is found to be 'b'.

Given that two values of distances measured represent the focal length values in the two cases, obtain a formula for the refractive index of the liquid.

If r = 10cm, a = 15cm, b = 10cm, find the refractive index of the liquid.

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Solution

The focal length $(f_1)$ of lens is given by

$\frac{1}{\text{f}_1}=(\text{n}-1)(\frac{1}{\text{r}}+\frac{1}{\text{r}})=\frac{2(\text{n}-1)}{\text{r}}$
Given $f_1= b$
$\Rightarrow\frac{1}{\text{b}}=\frac{2(\text{n}-1)}{\text{r}}$
$\Rightarrow\text{b}=\frac{\text{r}}{2(\text{n}-1)}$
The focal lenght of liquid lens (plano concave lens) is
$\frac{1}{\text{f}_2}=(\text{n}_{\text{l}}-1)(-\frac{1}{\text{r}}-\frac{1}{\propto})$
$=-\frac{(\text{n}_{\text{l}}-1)}{\text{r}}$
$\Rightarrow\text{f}_2=\frac{\text{r}}{(\text{n}_{\text{l}}-1)}$
As glass lens and liquid lens are in contact
$\therefore\frac{1}{\text{f}}=\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}=\frac{1}{\text{b}}-\frac{(\text{n}_{\text{l}}-1)}{\text{r}}$
Given $f = a$
$\therefore\frac{1}{\text{a}}=\frac{1}{\text{b}}-\frac{\text{n}_{\text{l}}-1}{\text{r}}$
$\Rightarrow\frac{\text{n}_{\text{l}}-1}{\text{r}}=\frac{1}{\text{b}}-\frac{1}{\text{a}}$
$\Rightarrow\text{n}_{\text{l}}-1=\text{r}\big(\frac{1}{\text{b}}-\frac{1}{\text{a}}\big)$
Refractive index of liquid,

$\text{n}_{\text{l}}=1+\frac{\text{r}}{\text{b}}-\frac{\text{r}}{\text{a}}$

$\text{n}_{\text{l}}=1+\frac{10}{10}-\frac{10}{15}$
$=1+1-\frac{2}{3}=2-\frac{2}{3}=\frac{4}{3}=1.33$

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