Question

The focal length of a convex lens of glass \[\left( {n = 1.50} \right)\] in the air is \[40cm\] . Calculate the focal length of the lens when it is immersed in a liquid of refractive index
(i) \[1.30\;\]
(ii) \[1.50\;\]
(iii) \[1.70\]

Answer 1

Hint: We can see from the lens maker's formula that the focal length of the lens is inversely proportional to the refractive index of the lens medium. As a result, when a lens is dipped in a denser substance than air, the net refractive index of the lens lowers, and the focal length increases.
A system with a shorter focal length bends the rays more sharply, bringing them closer to focus or diverging them faster.
The index of refraction of the glass, the radii of curvature of the surfaces, and the medium in which the lens lives determine the principal focal length of a lens.

Complete step-by-step solution:
Given,
\[{n_{glass{\text{ }}}} = 1.50\]
\[{f_{air}} = 40cm\]
The formula for calculating the focal length,
\[\dfrac{1}{f} = (\dfrac{{{n_2}}}{{{n_1}}} - 1)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\]
In which
\[f\]= focal length
\[{n_2},{n_1}\] = refractive indexes of the two mediums.
${R_1},{R_2}$ = Radius of curvature of the surface
\[ \Rightarrow \dfrac{1}{{{f_{air}}}} = (\dfrac{{{n_{glass}}}}{{{n_{air}}}} - 1)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\]
\[\dfrac{1}{{{f_{air}}}} = (\dfrac{{1.50}}{1} - 1)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\] \[[\because {n_{glass{\text{ }}}} = 1.50,{n_{air}} = 1]\]
Now,
\[\dfrac{1}{{{f_1}}} = \left( {\dfrac{{1.50}}{{1.30}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\] \[[\because {n_{glass{\text{ }}}} = 1.50,{n_1} = 1.30]\] (eq. 1)
\[\dfrac{1}{{{f_2}}} = \left( {\dfrac{{1.50}}{{1.50}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\] \[[\because {n_{glass{\text{ }}}} = 1.50,{n_2} = 1.50]\] (eq. 2)
\[\dfrac{1}{{{f_3}}} = \left( {\dfrac{{1.50}}{{1.70}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\] \[[\because {n_{glass{\text{ }}}} = 1.50,{n_3} = 1.50]\] (eq. 3)
Now, for \[{f_1}\],
\[\dfrac{{{f_1}}}{{{f_{air}}}} = \dfrac{{\left( {1.50 - 1} \right)}}{{\left( {\dfrac{{1.5}}{{1.3}} - 1} \right)}}\]
Since, \[{f_{air}} = 40cm\]
\[\dfrac{{{f_1}}}{{40}} = \dfrac{{\left( {.50} \right)}}{{\left( {\dfrac{{1.5 - 1.3}}{{1.3}}} \right)}}\]
\[ \Rightarrow {f_1} = \dfrac{{40 \times 0.5 \times 1.3}}{{0.2}}\]
\[ \Rightarrow {f_1} = 100 \times 1.3\]
\[ \Rightarrow {f_1} = 130cm\]
 for\[{f_2}\] ,
\[\dfrac{{{f_2}}}{{{f_{air}}}} = \dfrac{{\left( {1.50 - 1} \right)}}{{\left( {\dfrac{{1.50}}{{1.50}} - 1} \right)}}\]
\[ \Rightarrow \dfrac{{{f_2}}}{{40}} = \dfrac{{0.5}}{{1 - 1}}\]
\[ \Rightarrow {f_2} = \infty \]
for\[{\text{ }}{f_3}\] ,
\[\dfrac{{{f_3}}}{{{f_{air}}}} = \dfrac{{\left( {1.50 - 1} \right)}}{{\left( {\dfrac{{1.50}}{{1.70}} - 1} \right)}}\]
\[ \Rightarrow \dfrac{{{f_3}}}{{40}} = \dfrac{{0.5}}{{( - 0.2)}} \times 1.70\]
\[ \Rightarrow \dfrac{{{f_3}}}{{40}} = - \dfrac{5}{2} \times 1.70\]
\[ \Rightarrow {f_3} = - 100 \times 1.70\]
\[ \Rightarrow {f_3} = - 170cm\]
Hence, the lens will behave like a convex lens.

Note: A positive focal length system converges light, whereas a negative focal length system diverges light.
When light passes through a material with a higher refractive index, the angle of refraction is smaller than the angle of incidence, and the light is refracted toward the surface's normal.
The higher the refractive index, the closer light will travel in the normal direction. When light passes through a medium having a lower refractive index, it is refracted away from the normal and towards the surface.
When a lens or glass slab is immersed in a liquid with the same refractive index as the lens, the refractive index of the lens equals \[1\] with the liquid, When a lens or glass slab is immersed in a liquid with the same refractive index as the lens, the refractive index of the lens equals \[1\] concerning the liquid.