
A thin equiconvex glass lens ${\mu _g} = 1.5$ is being placed on the top of a vessel of height $h = 20\,cm$ as shown in fig. A luminous point source is being placed at the bottom of the vessel on the principal axis of the lens. When the air is on both sides of the lens the image of the luminous source is formed at a distance of $20\,cm$ from the lens outside the vessel. When the air inside the vessel is being replaced by a liquid of refractive index ${\mu _1}$, the image of the same source is being formed at a distance $30\,cm$ from the lens outside the vessel. If ${\mu _1}$ is approximately $\dfrac{x}{{100}}$. Find $x$?

Answer
123k+ views
Hint: The value of the $x$ can be determined by using the formula of the thin lens equation or the lens makers formula. And by equating the lens maker formula with the focal length formula, the radius of the curvature is determined. By using this radius of the curvature and using this radius of curvature value, the ${\mu _1}$ is determined.
Formula used:
The focal length of the lens is given by,
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Where, $f$ is the focal length of the lens, $v$ is the distance of the image from the lens and $u$ is the distance of the object from the lens.
The lens maker formula is given by,
$\dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Where, $f$ is the focal length of the lens, $\mu $ is the refractive index of the glass, ${R_1}$ and ${R_2}$ are the radius of the curvature.
Complete step by step solution:
Given that,
The refractive index of the glass is, ${\mu _g} = 1.5$,
The height of the vessel or distance of the object is, $h = u = 20\,cm$,
The distance of the image formed is, $v = 20\,cm$,
After some changes, the distance of the image formed is, $v = 30\,cm$.
Now,
The focal length of the lens is given by,
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\,......................\left( 1 \right)$
The lens maker formula is given by,
$\dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\,................\left( 2 \right)$
By equating the equation (1) and equation (2), then
$\dfrac{1}{v} - \dfrac{1}{u} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\,................\left( 3 \right)$
By substituting the refractive index of the glass, the distance of the object and the distance of the image formed in the above equation, then
$\dfrac{1}{{20}} - \dfrac{1}{{ - 20}} = \left( {1.5 - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Assume that the ${R_1}$ and ${R_2}$ are equal to $R$, then
$\dfrac{1}{{20}} - \dfrac{1}{{ - 20}} = \left( {1.5 - 1} \right)\left( {\dfrac{1}{R} - \dfrac{1}{{ - R}}} \right)$
On further simplification in the above equation, then
$\dfrac{1}{{20}} + \dfrac{1}{{20}} = \left( {1.5 - 1} \right)\left( {\dfrac{1}{R} + \dfrac{1}{R}} \right)$
By adding the terms in the above equation, then
$\dfrac{2}{{20}} = 0.5 \times \dfrac{2}{R}$
On further simplification in the above equation, then
\[\dfrac{1}{{10}} = \dfrac{1}{R}\]
By taking reciprocal in the above equation, then the above equation is written as,
$R = 10\,cm$
For the second case, when the vessel is being filled with the liquid, then the equation (3) is written as,
$\dfrac{{{\mu _{air}}}}{v} - \dfrac{{{\mu _1}}}{u} = \left( {\dfrac{{{\mu _g} - {\mu _1}}}{{{R_1}}} - \dfrac{{{\mu _{air}} - {\mu _g}}}{{{R_2}}}} \right)\,................\left( 4 \right)$
Assume that the ${R_1}$ and ${R_2}$ are equal to $R$, then
$\dfrac{{{\mu _{air}}}}{v} - \dfrac{{{\mu _1}}}{{ - u}} = \left( {\dfrac{{{\mu _g} - {\mu _1}}}{R} + \dfrac{{{\mu _{air}} - {\mu _g}}}{{ - R}}} \right)$
On further simplification in the above equation, then
$\dfrac{{{\mu _{air}}}}{v} + \dfrac{{{\mu _1}}}{u} = \left( {\dfrac{{{\mu _g} - {\mu _1}}}{R} - \dfrac{{{\mu _{air}} - {\mu _g}}}{R}} \right)$
By substituting the ${\mu _{air}}$, ${\mu _g}$, $v$, $u$ and $R$ values in the above equation, then
$\dfrac{1}{{30}} + \dfrac{{{\mu _1}}}{{20}} = \left( {\dfrac{{1.5 - {\mu _1}}}{{10}} - \dfrac{{1 - 1.5}}{{10}}} \right)$
On further simplification in the above equation, then
$\dfrac{1}{{30}} + \dfrac{{{\mu _1}}}{{20}} = \dfrac{{1.5}}{{10}} - \dfrac{{{\mu _1}}}{{10}} + \dfrac{{0.5}}{{10}}$
By rearranging the terms in the above equation, then
$\dfrac{{{\mu _1}}}{{20}} + \dfrac{{{\mu _1}}}{{10}} = \dfrac{{1.5}}{{10}} + \dfrac{{0.5}}{{10}} - \dfrac{1}{{30}}$
On further simplification in the above equation, then
$\dfrac{{30{\mu _1}}}{{200}} = \dfrac{2}{{10}} - \dfrac{1}{{30}}$
By cancelling the terms in the above equation, then
$\dfrac{{3{\mu _1}}}{{20}} = \dfrac{2}{{10}} - \dfrac{1}{{30}}$
By cross multiplying the terms in the above equation, then
$\dfrac{{3{\mu _1}}}{{20}} = \dfrac{{60 - 10}}{{300}}$
On further simplification in the above equation, then
$\dfrac{{3{\mu _1}}}{{20}} = \dfrac{{50}}{{300}}$
By cancelling the terms in the above equation, then
$\dfrac{{3{\mu _1}}}{{20}} = \dfrac{1}{6}$
By cross multiplying the terms in the above equation, then
${\mu _1} = \dfrac{{20}}{{18}}$
By cancelling the terms in the above equation, then
${\mu _1} = 1.11$
From the equation given in the question,
${\mu _1} = \dfrac{x}{{100}}$
By equating the terms, then
$1.11 = \dfrac{x}{{100}}$
From the above equation, the value of the $x$ is $111$.
Note: The values of the, $v$ the distance of the image from the lens is given as two values, the value of the , $v$ the distance of the image from the lens is given as $20\,cm$ in the first condition, then some changes are made in the construction, then the value of the, $v$ the distance of the image from the lens is given as $30\,cm$.
Formula used:
The focal length of the lens is given by,
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Where, $f$ is the focal length of the lens, $v$ is the distance of the image from the lens and $u$ is the distance of the object from the lens.
The lens maker formula is given by,
$\dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Where, $f$ is the focal length of the lens, $\mu $ is the refractive index of the glass, ${R_1}$ and ${R_2}$ are the radius of the curvature.
Complete step by step solution:
Given that,
The refractive index of the glass is, ${\mu _g} = 1.5$,
The height of the vessel or distance of the object is, $h = u = 20\,cm$,
The distance of the image formed is, $v = 20\,cm$,
After some changes, the distance of the image formed is, $v = 30\,cm$.
Now,
The focal length of the lens is given by,
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\,......................\left( 1 \right)$
The lens maker formula is given by,
$\dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\,................\left( 2 \right)$
By equating the equation (1) and equation (2), then
$\dfrac{1}{v} - \dfrac{1}{u} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\,................\left( 3 \right)$
By substituting the refractive index of the glass, the distance of the object and the distance of the image formed in the above equation, then
$\dfrac{1}{{20}} - \dfrac{1}{{ - 20}} = \left( {1.5 - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Assume that the ${R_1}$ and ${R_2}$ are equal to $R$, then
$\dfrac{1}{{20}} - \dfrac{1}{{ - 20}} = \left( {1.5 - 1} \right)\left( {\dfrac{1}{R} - \dfrac{1}{{ - R}}} \right)$
On further simplification in the above equation, then
$\dfrac{1}{{20}} + \dfrac{1}{{20}} = \left( {1.5 - 1} \right)\left( {\dfrac{1}{R} + \dfrac{1}{R}} \right)$
By adding the terms in the above equation, then
$\dfrac{2}{{20}} = 0.5 \times \dfrac{2}{R}$
On further simplification in the above equation, then
\[\dfrac{1}{{10}} = \dfrac{1}{R}\]
By taking reciprocal in the above equation, then the above equation is written as,
$R = 10\,cm$
For the second case, when the vessel is being filled with the liquid, then the equation (3) is written as,
$\dfrac{{{\mu _{air}}}}{v} - \dfrac{{{\mu _1}}}{u} = \left( {\dfrac{{{\mu _g} - {\mu _1}}}{{{R_1}}} - \dfrac{{{\mu _{air}} - {\mu _g}}}{{{R_2}}}} \right)\,................\left( 4 \right)$
Assume that the ${R_1}$ and ${R_2}$ are equal to $R$, then
$\dfrac{{{\mu _{air}}}}{v} - \dfrac{{{\mu _1}}}{{ - u}} = \left( {\dfrac{{{\mu _g} - {\mu _1}}}{R} + \dfrac{{{\mu _{air}} - {\mu _g}}}{{ - R}}} \right)$
On further simplification in the above equation, then
$\dfrac{{{\mu _{air}}}}{v} + \dfrac{{{\mu _1}}}{u} = \left( {\dfrac{{{\mu _g} - {\mu _1}}}{R} - \dfrac{{{\mu _{air}} - {\mu _g}}}{R}} \right)$
By substituting the ${\mu _{air}}$, ${\mu _g}$, $v$, $u$ and $R$ values in the above equation, then
$\dfrac{1}{{30}} + \dfrac{{{\mu _1}}}{{20}} = \left( {\dfrac{{1.5 - {\mu _1}}}{{10}} - \dfrac{{1 - 1.5}}{{10}}} \right)$
On further simplification in the above equation, then
$\dfrac{1}{{30}} + \dfrac{{{\mu _1}}}{{20}} = \dfrac{{1.5}}{{10}} - \dfrac{{{\mu _1}}}{{10}} + \dfrac{{0.5}}{{10}}$
By rearranging the terms in the above equation, then
$\dfrac{{{\mu _1}}}{{20}} + \dfrac{{{\mu _1}}}{{10}} = \dfrac{{1.5}}{{10}} + \dfrac{{0.5}}{{10}} - \dfrac{1}{{30}}$
On further simplification in the above equation, then
$\dfrac{{30{\mu _1}}}{{200}} = \dfrac{2}{{10}} - \dfrac{1}{{30}}$
By cancelling the terms in the above equation, then
$\dfrac{{3{\mu _1}}}{{20}} = \dfrac{2}{{10}} - \dfrac{1}{{30}}$
By cross multiplying the terms in the above equation, then
$\dfrac{{3{\mu _1}}}{{20}} = \dfrac{{60 - 10}}{{300}}$
On further simplification in the above equation, then
$\dfrac{{3{\mu _1}}}{{20}} = \dfrac{{50}}{{300}}$
By cancelling the terms in the above equation, then
$\dfrac{{3{\mu _1}}}{{20}} = \dfrac{1}{6}$
By cross multiplying the terms in the above equation, then
${\mu _1} = \dfrac{{20}}{{18}}$
By cancelling the terms in the above equation, then
${\mu _1} = 1.11$
From the equation given in the question,
${\mu _1} = \dfrac{x}{{100}}$
By equating the terms, then
$1.11 = \dfrac{x}{{100}}$
From the above equation, the value of the $x$ is $111$.
Note: The values of the, $v$ the distance of the image from the lens is given as two values, the value of the , $v$ the distance of the image from the lens is given as $20\,cm$ in the first condition, then some changes are made in the construction, then the value of the, $v$ the distance of the image from the lens is given as $30\,cm$.
Recently Updated Pages
How to find Oxidation Number - Important Concepts for JEE

How Electromagnetic Waves are Formed - Important Concepts for JEE

Electrical Resistance - Important Concepts and Tips for JEE

Average Atomic Mass - Important Concepts and Tips for JEE

Chemical Equation - Important Concepts and Tips for JEE

Concept of CP and CV of Gas - Important Concepts and Tips for JEE

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

JEE Main Login 2045: Step-by-Step Instructions and Details

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Charging and Discharging of Capacitor

Physics Average Value and RMS Value JEE Main 2025

Degree of Dissociation and Its Formula With Solved Example for JEE
