Question

A thin equiconvex glass lens of refractive index $1.5$ has power of $5D$. When the lens is immersed in a liquid of refractive index $\mu$ , it acts as a diverging lens of focal length $100cm$ . THE value of $\mu$ of liquid is
(A) $\dfrac{4}{3}$
(B) $\dfrac{3}{4}$
(C) $\dfrac{{15}}{{11}}$
(D) $\dfrac{8}{3}$

Answer 1

Hint
${}^x{\mu _y}$ tells that refractive index of $y$ with respect to $x$. Use the formula of power of lens and focal length. After that use the formula of power by using lens maker’s formula which is $P = (\mu - 1)[\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}]$.

Complete step by step solution
Here , Power Is given by , $P = + 5{\text{ D}}$,
Also Refractive index of glass, ${}^a{\mu _g}$ is $1.5$.
When the lens in air,
${f_a} = \dfrac{1}{P} = \dfrac{1}{5}m = 20cm$ , ${f_a}$ is focal length of lens in air
But we know that $\dfrac{1}{{{f_a}}} = ({}^a{\mu _g} - 1)[\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}]$ ----------------(i)
where ${}^a{\mu _g}$ is a refractive index of glass with respect to air and ${R_1} \& {R_2}$ are the radii of curvature.
Now putting above values in equation (i);
We get, $\dfrac{1}{{20}} = (1.5 - 1)[\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}]$
On further solving, $\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}} = \dfrac{1}{{20 \times 0.5}} = \dfrac{1}{{10}}$ -------------(ii)
Now when the lens is immersed in liquid,
$\dfrac{1}{{{f_l}}} = (\dfrac{{{}^a{\mu _g}}}{{{}^a{\mu _l}}} - 1)[\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}]$ --------------------(iii)
Where ${f_l}$ is the focal length of lens in liquid, ${}^a{\mu _l}$is refractive index of liquid with respect to air and ${R_1}\& {R_2}$ are the radii of curvature.
Now putting all values in equation (iii);
We get, $\dfrac{1}{{100}} = (\dfrac{{1.5}}{{{}^a{\mu _l}}} - 1) \times \dfrac{1}{{10}}$
On further solving, $\dfrac{{1.5}}{{{}^a{\mu _l}}} - 1 = - \dfrac{1}{{10}}$
On more simplifying, $\dfrac{{1.5}}{{{}^a{\mu _l}}} = 1 + 0.1 = 1.1$
Therefore, ${}^a{\mu _l} = \dfrac{{15}}{{11}}$.
Hence we get the refractive index of liquid which is $\dfrac{{15}}{{11}}$ .
So the correct answer is option (C).

Note
Remember Power of lens is the reciprocal of focal length of lens in metre or Power of lens is hundred times the reciprocal of focal length of lens in centimetre. Also focal length of converging lens is positive and that of diverging lens is negative.