Answer

Verified

51.9k+ views

**Hint:**We can solve this question by using the formula of focal length in terms of refractive index. By using the given value of the refractive index of glass and the focal length we will evaluate the differences between the radius of curvature as a constant term. Then by using that constant and substituting we will find the value of the focal length of the lens when it will be immersed in water.

Formula used:

Focal length formula

$\dfrac{1}{f} = \left( {\mu - 1} \right)\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right]$

where, $\mu $ is the refractive index, ${R_1}$and ${R_2}$ is the radius of curvature.

**Complete step-by-step solution:**

For the convex glass having refractive index given as $\left( {{\mu _g} = 1.5} \right)$, we will substitute it in the formula of focal length such as

$\dfrac{1}{f} = \left( {{\mu _g} - 1} \right)\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right]$

Now substituting the value of focal length which is given as $8cm$ for the convex glass and the value of ${\mu _g} = 1.5$, hence

$\dfrac{1}{8} = \left( {1.5 - 1} \right)\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right]$

$ \Rightarrow \dfrac{1}{8} = \left( {0.5} \right)\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right]$

Hence using the transposition method arranging the terms we will get

$\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right] = \dfrac{1}{4}$ ………. $(1)$

As the value of the radius of the curvature term is constant hence it will remain as it is and this value can be used further.

Now we will consider the same formula of focal length for the case when the glass is immersed in water. Hence the relative refractive index is glass and water is given as

$\dfrac{{{\mu _g}}}{{{\mu _w}}} = \dfrac{{1.5}}{{\dfrac{4}{3}}} \approx \dfrac{{1.5}}{{1.33}}$

The focal length formula can be given as

$\dfrac{1}{{{f_w}}} = \left( {\dfrac{{{\mu _g}}}{{{\mu _w}}} - 1} \right)\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right]$

$ \Rightarrow \dfrac{1}{{{f_w}}} = \left( {\dfrac{{1.5}}{{1.33}} - 1} \right)\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right]$

Substituting the value of constant term from the equation $(1)$, hence

$\dfrac{1}{{{f_w}}} = \left( {\dfrac{{1.5}}{{1.33}} - 1} \right) \times \dfrac{1}{4}$

$ \Rightarrow \dfrac{1}{{{f_w}}} = \dfrac{1}{{32}}$

$\therefore {f_w} = 32cm$

**The focal length lens when it is immersed in water is given as ${f_w} = 32cm$.**

**Note:**Here we have used the formula focal length in terms of refractive index, where the refractive index can be measured as the bending of a light ray when it passes through one medium to another. It can also be defined as the ratio of the velocity of light in a vacuum to the velocity of light in a given substance.

Recently Updated Pages

Which is not the correct advantage of parallel combination class 10 physics JEE_Main

State two factors upon which the heat absorbed by a class 10 physics JEE_Main

What will be the halflife of a first order reaction class 12 chemistry JEE_Main

Which of the following amino acids is an essential class 12 chemistry JEE_Main

Which of the following is least basic A B C D class 12 chemistry JEE_Main

Out of the following hybrid orbitals the one which class 12 chemistry JEE_Main

Other Pages

The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main

Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main

Which of the following sets of displacements might class 11 physics JEE_Main

Explain the construction and working of a GeigerMuller class 12 physics JEE_Main