Question

A thin equiconvex lens is made of a glass of refractive index 1.5 and its focal length is 0.2m. If it acts as a concave lens of 0.5m focal length when dipped in a liquid, the refractive index of the liquid is:
$
  (a){\text{ }}\dfrac{{17}}{8} \\
  (b){\text{ }}\dfrac{{15}}{8} \\
  (c){\text{ }}\dfrac{{13}}{8} \\
  (d){\text{ }}\dfrac{9}{8} \\
 $

Answer 1

Hint: In this question use the direct relationship between focal length and the refractive index which is $\dfrac{1}{f} = \left( {\dfrac{{{\mu _1}}}{{{\mu _2}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$. When the lens is made up of glass it is simply present in air so use ${\mu _1} = {\mu _g},{\mu _2} = {\mu _{air}}$. Now when lens is dipped in liquid ${\mu _1} = {\mu _g},{\mu _2} = {\mu _L}$where ${\mu _L}$ is the refractive index of the liquid. Use these concepts to get the value of the refractive index of liquid.

Complete Step-by-Step solution:
Given data:
Refractive index of glass = ${\mu _g} = 1.5$
Focal length of convex lens = ${f_1} = 0.2$ m.
Focal length of concave lens = ${f_2} = - 0.5$m (as focal length of concave lens is always negative).
Now as we know the relation between focal length and refractive index is
$ \Rightarrow \dfrac{1}{f} = \left( {\dfrac{{{\mu _1}}}{{{\mu _2}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
So when it behaves as a convex lens made up of glass so the first medium is glass and the second medium is air.
Therefore, $f = {f_1},{\mu _1} = {\mu _g},{\mu _2} = {\mu _{air}} = 1$ (as refractive index of air is 1)
Now substitute the values we have,
$ \Rightarrow \dfrac{1}{{0.2}} = \left( {\dfrac{{1.5}}{1} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$...................... (1)
Now when it behaves as a concave lens and dipped into the liquid so the first medium is glass and the second medium is liquid.
Therefore,$f = {f_2},{\mu _1} = {\mu _g},{\mu _2} = {\mu _L}$, where ${\mu _L}$ is the refractive index of the liquid.
Now substitute the values we have,
$ \Rightarrow \dfrac{1}{{ - 0.5}} = \left( {\dfrac{{1.5}}{{{\mu _L}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$..................... (2)
Now divide equation (1) and (2) we have,
$ \Rightarrow \dfrac{{\dfrac{1}{{0.2}}}}{{\dfrac{1}{{ - 0.5}}}} = \dfrac{{\left( {\dfrac{{1.5}}{1} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)}}{{\left( {\dfrac{{1.5}}{{{\mu _L}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)}}$
Now simplify the above equation we have,
$ \Rightarrow - \dfrac{5}{2} = \dfrac{{\left( {\dfrac{{1.5}}{1} - 1} \right)}}{{\left( {\dfrac{{1.5}}{{{\mu _L}}} - 1} \right)}}$
\[ \Rightarrow 5\left( {\dfrac{{1.5}}{{{\mu _L}}} - 1} \right) = - 2\left( {0.5} \right) = - 1\]
$ \Rightarrow \dfrac{{7.5}}{{{\mu _L}}} = - 1 + 5 = 4$
$ \Rightarrow {\mu _L} = \dfrac{{7.5}}{4} = \dfrac{{75}}{{4 \times 10}} = \dfrac{{15}}{8}$
So this is the required answer.
Hence option (B) is the correct answer.

Note – There is a trick to eliminate options while solving problems of these kinds in which a convex lens behaves as a concave lens and vice versa when dipped in other liquid. It is only possible that a convex or a concave lens changes its behavior after dipping in some liquid if the refractive index of that liquid is greater than the refractive index of the lens. So in this case any option whose value is less than 1.5 is wrong.