Question

A thin liquid convex lens is formed in the glass. Refractive index of the liquid is \[\dfrac{4}{3}\] and that of the glass is \[\dfrac{3}{2}\]. If ‘f’ is the focal length of the liquid lens in air, its focal length and nature in the glass are:
A. f, convex
B. f, concave
C. 2f, concave
D. 3f, concave

Answer 1

Hint:Use lens maker’s formula to determine the radius of curvature of the liquid convex lens in terms of its focal length in the air medium. Use the same formula to determine the focal length of the liquid lens in terms of its focal length in the air. If the focal length is negative, the lens should be the concave lens.

Formula used:
\[\dfrac{1}{f} = \left( {\dfrac{{{\mu _l}}}{{{\mu _a}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}} \right)\]
Here, f is the focal length of the liquid convex lens, \[{\mu _l}\] is the refractive index of the liquid convex lens, \[{\mu _a}\] is the refractive index of the air, \[{R_1}\] and \[{R_2}\] are the radii of curvatures of the two surfaces of the convex lens.

Complete step by step answer:
We know that when we place a convex lens in the medium of different refractive, the convex lens behaves differently. We have given that the convex lens is placed in the glass and the refractive index \[\dfrac{3}{2}\]. Let us calculate the focal length of the liquid convex lens when it is placed in the air medium using the lens maker’s formula,
\[\dfrac{1}{f} = \left( {\dfrac{{{\mu _l}}}{{{\mu _a}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}} \right)\]
Here, f is the focal length of the liquid convex lens, \[{\mu _l}\] is the refractive index of the liquid convex lens, \[{\mu _a}\] is the refractive index of the air, \[{R_1}\] and \[{R_2}\] are the radii of curvatures of the two surfaces of the convex lens.

We know that the convex lens is symmetric and therefore, the radii of curvatures of the two surfaces of the lens are the same. Thus, we can write the above equation as,
\[\dfrac{1}{f} = \left( {\dfrac{{{\mu _l}}}{{{\mu _a}}} - 1} \right)\left( {\dfrac{2}{R}} \right)\]
Substituting \[\dfrac{4}{3}\] for \[{\mu _l}\] and 1 for \[{\mu _a}\] in the above equation, we get,
\[\dfrac{1}{f} = \left( {\dfrac{{4/3}}{1} - 1} \right)\left( {\dfrac{2}{R}} \right)\]
\[ \Rightarrow \dfrac{3}{f} = \left( {\dfrac{2}{R}} \right)\] …… (1)
Let us calculate the focal length of the liquid convex lens when it is placed in the glass using the lens maker’s formula,
\[\dfrac{1}{{{f_l}}} = \left( {\dfrac{{{\mu _l}}}{{{\mu _g}}} - 1} \right)\left( {\dfrac{2}{R}} \right)\]
Here, \[{\mu _g}\] is the refractive index of the glass and \[{f_l}\] is the focal length of the liquid convex lens.

Substituting \[\dfrac{4}{3}\] for \[{\mu _l}\] and \[\dfrac{3}{2}\] for \[{\mu _g}\] in the above equation, we get,
\[\dfrac{1}{{{f_l}}} = \left( {\dfrac{{4/3}}{{3/2}} - 1} \right)\left( {\dfrac{2}{R}} \right)\]
\[ \Rightarrow \dfrac{1}{{{f_l}}} = \left( {\dfrac{8}{9} - 1} \right)\left( {\dfrac{2}{R}} \right)\]
\[ \Rightarrow \dfrac{1}{{{f_l}}} = \left( { - \dfrac{1}{9}} \right)\left( {\dfrac{2}{R}} \right)\]
Using equation (1) in the above equation, we get,
\[\dfrac{1}{{{f_l}}} = \left( { - \dfrac{1}{9}} \right)\left( {\dfrac{3}{f}} \right)\]
\[ \Rightarrow \dfrac{1}{{{f_l}}} = - \dfrac{1}{{3f}}\]
\[ \therefore {f_l} = - 3f\]
The negative sign in the above expression shows the convex lens will now behave as the concave lens since the focal length of the concave lens is negative.The focal length of the liquid lens in the glass is 3f.

So, the correct answer is option D.

Note:Students must know the signs of focal length of convex lens and concave lens in order to answer these types of questions. In the solution, we have derived the lens maker’s formula for the convex lens since the convex lens has the symmetric surface on both sides. If we have the Plano-convex lens, the lens maker’s formula would have two different radii of curvatures.