Question

A thin lens is made with a material having refractive index \[\mu = 1.5\]. Both the sides are convex. It is dipped in water ( \[\mu = 1.33\] ), it will behave like:
A. A convergent lens
B. A divergent lens
C. A rectangular slab
D. A prism

Answer 1

Hint:First of all, we will use the lens maker’s formula to find the focus of the lens in air. After that we will again use the lens maker’s formula, after we have immersed the lens in water. We will try to find the focal length in water in terms of focal length in air. After that we will compare the signs if they are the same or different to the original one.

Formula used:
Using lens maker’s equation
\[\dfrac{1}{f} = \left( {n - 1} \right)\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}} + \dfrac{{\left( {n - 1} \right)d}}{{\left( {n{R_1}{R_2}} \right)}}} \right]\] …… (1)
Where,
\[f\] is the lens focal length
\[n\] is refractive index
\[{R_1}\] is the radius of curvature of the lens which is closest to the light source.
\[{R_2}\] is the radius of curvature of the lens which is farthest to the light source.
\[d\] is the thickness of the lens.

Complete step by step answer:
In the given question, we are supplied with the following data:
The refractive index of the material is \[1.33\]. The refractive index of the medium is \[1.55\].
We know that for thin glasses \[d\]can be ignored so we get,
\[\dfrac{1}{f} = \left( {n - 1} \right)\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right]\].

Therefore,
Let, the focal length of the lens in air be \[{f_1}\]. So we get,
$\dfrac{1}{{{f_1}}} = \left( {n - 1} \right)\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right] \\
\Rightarrow \left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right] = \dfrac{1}{{{f_1}\left( {n - 1} \right)}} \\
\Rightarrow \left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right] = \dfrac{1}{{{f_1}\left( {1.5 - 1} \right)}} \\
\Rightarrow \left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right] = \dfrac{2}{{{f_1}}} \\$
Now, when the lens is dipped into a refractive index medium \[1.33\], so \[{f_2}\] be the focal length of lens when dipped in water medium, so we get,
$\dfrac{1}{{{f_2}}} = \left( {n - 1} \right)\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right] \\
\Rightarrow \dfrac{1}{{{f_2}}} = \left( {\dfrac{n}{{n' - 1}}} \right)\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right] \\
\Rightarrow \dfrac{1}{{{f_2}}} = \left( {\dfrac{{1.5}}{{1.33 - 1}}} \right) \times \dfrac{2}{{{f_1}}} \\
\Rightarrow \dfrac{1}{{f2}} = \dfrac{{0.34}}{{1.33\left( {{f_1}} \right)}} \\$
\[ \therefore {f_2} = 3.91\left( {{f_1}} \right)\]
Hence, the sign of \[{f_2}\] is the same as that of \[{f_1}\], so the lens will behave as a converging lens.

The correct option is A.

Additional information:
Lens maker’s formula: The formula of the lens maker is the relationship between the focal length of a lens to its material's refractive index and the curvature radii of its two surfaces. It is used by manufacturers of lenses to make special power lenses from the glass of a given refractive index.

Note: It is important to remember that the focus of a converging lens is always positive, while the focus of a diverging lens is always negative. It is where most of the students tend to make mistakes. When immersed in water, the focal length of a lens is greatly influenced. The explanation for this is that the angle of refraction depends on the relative variation between the refraction indices of the incoming medium and the lens medium.