Curve fitting is the process of constructing a curve or mathematical function, that has the best fit to a series of data points. Fitted curves can be used as an aid for data visualization, to conclude values of a function where no data are available, and to summarize the relationships among two or more variables.
Equation for Cauchy Curve is, $$ Y=\frac { I }{ \left[ 1+{ \left( \frac { x-a }{ \gamma } \right) }^{ 2 } \right] } $$ Where \(I\) is intensity,
\(a\) is Position,
\(\gamma\) is Scale parameter

Curve fitting is the process of constructing a curve or mathematical function, that has the best fit to a series of data points. Fitted curves can be used as an aid for data visualization, to conclude values of a function where no data are available, and to summarize the relationships among two or more variables.
Equation for Cauchy Curve is, $$ Y=\frac { I }{ \left[ 1+{ \left( \frac { x-a }{ \gamma } \right) }^{ 2 } \right] } $$ Where \(I\) is intensity,
\(a\) is Position,
\(\gamma\) is Scale parameter

Curve fitting is the process of constructing a curve or mathematical function, that has the best fit to a series of data points. Fitted curves can be used as an aid for data visualization, to conclude values of a function where no data are available, and to summarize the relationships among two or more variables.
Equation for Gaussian Curve is, $$ Y = A.{ e }^{ -\left( \frac { { \left( x-B \right) }^{ 2 } }{ 2{ C }^{ 2 } } \right) } $$ Where \(A\) is Height,
\(B\) is Position and
\(C\) is FWHM.

Curve fitting is the process of constructing a curve or mathematical function, that has the best fit to a series of data points. Fitted curves can be used as an aid for data visualization, to conclude values of a function where no data are available, and to summarize the relationships among two or more variables.
Equation for Gaussian Curve is, $$ Y = A.{ e }^{ -\left( \frac { { \left( x-B \right) }^{ 2 } }{ 2{ C }^{ 2 } } \right) } $$ Where \(A\) is Height,
\(B\) is Position and
\(C\) is FWHM.

Curve fitting is the process of constructing a curve or mathematical function, that has the best fit to a series of data points. Fitted curves can be used as an aid for data visualization, to conclude values of a function where no data are available, and to summarize the relationships among two or more variables.
Equation for Exponential Decay Curve is, $$ Y = \frac { 1 }{ B } .{ e }^{ \left( \frac { { A-x } }{ B } \right) } $$Where \(A\) is Position,
\(B\) is Scale parameter

Curve fitting is the process of constructing a curve or mathematical function, that has the best fit to a series of data points. Fitted curves can be used as an aid for data visualization, to conclude values of a function where no data are available, and to summarize the relationships among two or more variables.
Equation for Exponential Decay Curve is, $$ Y = \frac { 1 }{ B } .{ e }^{ \left( \frac { { A-x } }{ B } \right) } $$Where \(A\) is Position,
\(B\) is Scale parameter

Curve fitting is the process of constructing a curve or mathematical function, that has the best fit to a series of data points. Fitted curves can be used as an aid for data visualization, to conclude values of a function where no data are available, and to summarize the relationships among two or more variables. An equation for Straight Line is, Y = MX + C Where M is slope and C is intercept

Curve fitting is the process of constructing a curve or mathematical function, that has the best fit to a series of data points. Fitted curves can be used as an aid for data visualization, to conclude values of a function where no data are available, and to summarize the relationships among two or more variables. An equation for Straight Line is, Y = MX + C Where M is slope and C is intercept

Plane Mirror

Image formed is as far behind the mirror as the object is in front of it, with the lateral inversion of image. i.e. right hand side of object is left hand side of image and vice versa.

Convex Mirror

Image formed is always virtual, erect and diminished.

Concave Mirror

Depending upon the position of the object, size of the image formed can be equal to, larger than or smaller than the size of object. Also the nature of image formed may be real or virtual.

Plane Mirror

Image formed is as far behind the mirror as the object is in front of it, with the lateral inversion of image. i.e. right hand side of object is left hand side of image and vice versa.

Convex Mirror

Image formed is always virtual, erect and diminished.

Concave Mirror

Depending upon the position of the object, size of the image formed can be equal to, larger than or smaller than the size of object. Also the nature of image formed may be real or virtual.

The distance between the principal focus of a mirror and the pole of mirror is called focal length of the mirror.

In case of Concave mirror distance between the pole ‘O’ and principal focus ‘F’ of mirror is focal length and is always taken as negative.

In case of Convex mirror distance between the pole ‘O’ and principal focus ‘F’ of mirror is focal length and is always taken as positive.

The radius of Curvature of plane mirror is infinite, hence the focal length of plane mirror will also be infinite.

The distance between the principal focus of a mirror and the pole of mirror is called focal length of the mirror.

In case of Concave mirror distance between the pole ‘O’ and principal focus ‘F’ of mirror is focal length and is always taken as negative.

In case of Convex mirror distance between the pole ‘O’ and principal focus ‘F’ of mirror is focal length and is always taken as positive.

The radius of Curvature of plane mirror is infinite, hence the focal length of plane mirror will also be infinite.

Solution: For a concave mirror u becomes negative, $$\therefore u = -7.5 cm $$ (i) When real image is formed v is also negative $$ \therefore v = -30 cm $$

By using mirror formula, $$ \frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } $$ Lets substitute the values, $$ \frac { 1 }{ -7.5 } +\frac { 1 }{ -30 } =\frac { 1 }{ f } $$ $$ \therefore \frac { 1 }{ f } = \frac { -30-7.5 }{ \left( -7.5 \right) \times \left( -30 \right) } = \frac { -37.5 }{ 225 } $$ $$ \therefore f = \frac { 225 }{ -37.5 } = -6 cm $$ (ii) When virtual image is formed v is positive $$ \therefore v = 30 cm $$ By using mirror formula, $$ \frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } $$ Lets substitute the values, $$ \frac { 1 }{ -7.5 } +\frac { 1 }{ 30 } =\frac { 1 }{ f } $$ $$ \therefore \frac { 1 }{ f } = \frac { 30-7.5 }{ \left( -7.5 \right) \times \left( 30 \right) } $$ $$= \frac { 22.5 }{ -225 } $$ $$ \therefore f = \frac { -225 }{ 22.5 } = -10 cm $$ Hence the focal length of mirror when real image is formed is -6 cm and when virtual image is formed at 30 cm, focal length of mirror will be -10 cm.

Solution: For a concave mirror u becomes negative, $$\therefore u = -7.5 cm $$ (i) When real image is formed v is also negative $$ \therefore v = -30 cm $$

By using mirror formula, $$ \frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } $$ Lets substitute the values, $$ \frac { 1 }{ -7.5 } +\frac { 1 }{ -30 } =\frac { 1 }{ f } $$ $$ \therefore \frac { 1 }{ f } = \frac { -30-7.5 }{ \left( -7.5 \right) \times \left( -30 \right) } = \frac { -37.5 }{ 225 } $$ $$ \therefore f = \frac { 225 }{ -37.5 } = -6 cm $$ (ii) When virtual image is formed v is positive $$ \therefore v = 30 cm $$ By using mirror formula, $$ \frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } $$ Lets substitute the values, $$ \frac { 1 }{ -7.5 } +\frac { 1 }{ 30 } =\frac { 1 }{ f } $$ $$ \therefore \frac { 1 }{ f } = \frac { 30-7.5 }{ \left( -7.5 \right) \times \left( 30 \right) } $$ $$= \frac { 22.5 }{ -225 } $$ $$ \therefore f = \frac { -225 }{ 22.5 } = -10 cm $$ Hence the focal length of mirror when real image is formed is -6 cm and when virtual image is formed at 30 cm, focal length of mirror will be -10 cm.

Solution: As we have a concave mirror focal length is negative, $$ \therefore f = -50 cm $$ and $$ Magnification (m) = \frac { v }{ u } = 2 $$ $$ \therefore v = 2u $$ By using mirror formula, $$ \frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } $$ Lets substitute the values, $$ \frac { 1 }{ u } +\frac { 1 }{ 2u } =\frac { 1 }{ -50 } \frac { 2+1 }{ 2u } $$ $$=\frac { -1 }{ 50 } \frac { 3 }{ 2u } $$ $$ =\frac { -1 }{ 50 } $$ $$ \therefore 2u = -\left( 3 \times 50 \right) $$ or $$ u = \frac { -150 }{ 2 } = -75 cm $$

Solution: As we have a concave mirror focal length is negative, $$ \therefore f = -50 cm $$ and $$ Magnification (m) = \frac { v }{ u } = 2 $$ $$ \therefore v = 2u $$ By using mirror formula, $$ \frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } $$ Lets substitute the values, $$ \frac { 1 }{ u } +\frac { 1 }{ 2u } =\frac { 1 }{ -50 } \frac { 2+1 }{ 2u } $$ $$=\frac { -1 }{ 50 } \frac { 3 }{ 2u } $$ $$ =\frac { -1 }{ 50 } $$ $$ \therefore 2u = -\left( 3 \times 50 \right) $$ or $$ u = \frac { -150 }{ 2 } = -75 cm $$

(i) The pole of mirror is taken as origin.

(ii) All distances are measured from the pole ‘O’ of the mirror.

(iii) Rays of light are assumed to be coming from left side and enter the right side.

(iv) Distances measured in the direction of incident ray are taken as positive.

(v) All distances measured in direction opposite to incident ray are taken as negative.

(vi) Height measured above the principal axis are taken as positive and below as negative.

(vii) For concave mirrors, radius of curvature (R) and focal length (f) are taken as negative.

(viii) For convex mirrors, radius of curvature (R) and focal length (f) are taken as positive.

(ix) For real image magnification is negative and for virtual image magnification is positive.

(i) The pole of mirror is taken as origin.

(ii) All distances are measured from the pole ‘O’ of the mirror.

(iii) Rays of light are assumed to be coming from left side and enter the right side.

(iv) Distances measured in the direction of incident ray are taken as positive.

(v) All distances measured in direction opposite to incident ray are taken as negative.

(vi) Height measured above the principal axis are taken as positive and below as negative.

(vii) For concave mirrors, radius of curvature (R) and focal length (f) are taken as negative.

(viii) For convex mirrors, radius of curvature (R) and focal length (f) are taken as positive.

(ix) For real image magnification is negative and for virtual image magnification is positive.

If two charges \({ q }_{ 1 }\) and \({ q }_{ 2 }\) are separated by distance \(r\) in vacuum then electrostatic force between them will be given by the Coulomb law as below:

$${ F }_{ v } = \frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \qquad ...(1)$$ Now if these charges are placed in a medium of absolute permittivity \(\epsilon\) , then electrostatic force between these charges will be $${ F }_{ m } = \frac { 1 }{ 4\pi \epsilon } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \qquad ...(2)$$ We know that $$\epsilon = { \epsilon }_{ 0 } { \epsilon }_{ r } =K{ \epsilon }_{ 0 }$$ where \(K\) is dielectric constant or Relative permittivity or specific inductive capacity. or $$K = \frac { \epsilon }{ { \epsilon }_{ 0 } }$$ Now, eqn. (2) becomes, $${ F }_{ m } = \frac { 1 }{ 4\pi K{ \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \qquad ...(3)$$ Dividing eqn. (1) by eqn. (3), $$\frac { { F }_{ v } }{ { F }_{ m } } = \frac { \left( \frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \right) }{ \left( \frac { 1 }{ 4\pi K{ \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \right) } = K$$ Hence, dielectric constant (or relative permittivity) of a medium is defined as the ratio of the electrostatic force between two point charges when placed a certain distance apart in vacuum to the electrostatic force between the same charges when placed same distance apart in a medium.

This means a dielectric reduces the force between two charges.

If two charges \({ q }_{ 1 }\) and \({ q }_{ 2 }\) are separated by distance \(r\) in vacuum then electrostatic force between them will be given by the Coulomb law as below:

$${ F }_{ v } = \frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \qquad ...(1)$$ Now if these charges are placed in a medium of absolute permittivity \(\epsilon\) , then electrostatic force between these charges will be $${ F }_{ m } = \frac { 1 }{ 4\pi \epsilon } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \qquad ...(2)$$ We know that $$\epsilon = { \epsilon }_{ 0 } { \epsilon }_{ r } =K{ \epsilon }_{ 0 }$$ where \(K\) is dielectric constant or Relative permittivity or specific inductive capacity. or $$K = \frac { \epsilon }{ { \epsilon }_{ 0 } }$$ Now, eqn. (2) becomes, $${ F }_{ m } = \frac { 1 }{ 4\pi K{ \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \qquad ...(3)$$ Dividing eqn. (1) by eqn. (3), $$\frac { { F }_{ v } }{ { F }_{ m } } = \frac { \left( \frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \right) }{ \left( \frac { 1 }{ 4\pi K{ \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \right) } = K$$ Hence, dielectric constant (or relative permittivity) of a medium is defined as the ratio of the electrostatic force between two point charges when placed a certain distance apart in vacuum to the electrostatic force between the same charges when placed same distance apart in a medium.

This means a dielectric reduces the force between two charges.

Permittivity is the property of a particular medium which affects the magnitude of the force existing between two point charges.

We know that the greater the value of the permittivity of the medium placed between the two charged bodies the lesser the value of force existing between them.

Absolute permittivity \({ \epsilon }_{ 0 }\)

Relative permittivity \({ \epsilon }_{ r }\)

The absolute permittivity of air or vacuum is minimum and its value is \(8.854 \times { 10 }^{ -12 }\) F/m (farad/metre) whereas the value of absolute or (actual) permittivity \( \epsilon \) of all other insulating medium is more than \({ \epsilon }_{ 0 }\).

The ratio of these two permittivities i.e. absolute permittivity (\( \epsilon \)) of the insulating medium to the absolute permittivity (\({ \epsilon }_{ 0 }\)) of the air or vacuum is known as relative permittivity of that medium and is denoted by \({ \epsilon }_{ r }\). i.e. $$\quad { \epsilon }_{ r } = \frac { \epsilon }{ { \epsilon }_{ 0 } } $$

Permittivity is the property of a particular medium which affects the magnitude of the force existing between two point charges.

We know that the greater the value of the permittivity of the medium placed between the two charged bodies the lesser the value of force existing between them.

Absolute permittivity \({ \epsilon }_{ 0 }\)

Relative permittivity \({ \epsilon }_{ r }\)

The absolute permittivity of air or vacuum is minimum and its value is \(8.854 \times { 10 }^{ -12 }\) F/m (farad/metre) whereas the value of absolute or (actual) permittivity \( \epsilon \) of all other insulating medium is more than \({ \epsilon }_{ 0 }\).

The ratio of these two permittivities i.e. absolute permittivity (\( \epsilon \)) of the insulating medium to the absolute permittivity (\({ \epsilon }_{ 0 }\)) of the air or vacuum is known as relative permittivity of that medium and is denoted by \({ \epsilon }_{ r }\). i.e. $$\quad { \epsilon }_{ r } = \frac { \epsilon }{ { \epsilon }_{ 0 } } $$

Coulomb performed a number of experiments to see the effect of placing two small charges near each other. On his experimental researches, he established a law which is known as Coulombs Law of Electrostatics.

According to Coulombs Law :The force of attraction or repulsion between two charges is directly proportional to the product of magnitude of the two charges and inversely proportional to the square of distance between them.

If two point charges \({ q }_{ 1 } \) and \({ q }_{ 2 }\) have distance d between them, then force F between the charges can be mathematically expressed as: $$ F \propto { q }_{ 1 }{ q }_{ 2 }\qquad ...(1)$$ $$ F \propto \frac { 1 }{ { d }^{ 2 } } \qquad ...(2)$$ Combining (1) and (2), $$ F \propto \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } or F = K\frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } \qquad ...(3) $$ Where, \(K\) is a constant of proportionality and its value depends upon the medium in which the charges are placed and the system of units used.

In SI units force is measured in newton, charge in coulomb, distance in metre and the value of \(K\) is given as, $$ K = \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \qquad ...(4) (in SI system)$$ where, \({ \epsilon }_{ 0 }\) = Absolute permittivity of vacuum or permittivity of free space = \(8.854 \times { 10 }^{ -12 }\) farad/metre
\( { \epsilon }_{ r }\) = Relative permittivity of the medium w.r.t. vacuum in which the charges are placed (e.g. for air \({ \epsilon }_{ r }\) = 1)

Putting the value of \(K\) in equation (3), we get, $$ \boxed { F = \left( \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \right) \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } } \qquad ... (5)$$ Unit Charge or one coulomb charge in SI system can be defined as the amount of charge which when placed at a distance of one metre from an equal and similar charge in air, is repelled with a force of \(9 \times { 10 }^{ 9 } \)newton from it.

As per this defination, $$ { q }_{ 1 } = { q }_{ 2 } = q, $$ $$F = 9 \times { 10 }^{ 9 } newton,$$ $$ { \epsilon }_{ 0 } = 8.854 \times { 10 }^{ -12 } farad/metre,$$ \({ \epsilon }_{ r } = 1\) (Assuming two charges to be placed in air)
d = 1m

Putting these values in the above relation, we get $$ 9 \times { 10 }^{ 9 } = \frac { q \times q }{ 4\pi \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 \times { 1 }^{ 2 } } $$ $$ 9 \times { 10 }^{ 9 } = \frac { { q }^{ 2 } }{ 4\pi \times \left( 8.854 \times { 10 }^{ -12 } \right) } $$ $$ { q }^{ 2 } = \left( 9 \times { 10 }^{ 9 } \right) \times 4\pi \times \left( 8.854 \times { 10 }^{ -12 } \right) = 1.00086 $$ $$ \therefore q = \pm 1.00043 \ coulomb $$ Hence, unit charge.

Determine the force between two charges, each of one coulomb when they are separated at one metre distance in air.

Solution:We have, $${ q }_{ 1 } = { q }_{ 2 } = 1 \ coulomb $$ $$ { \epsilon }_{ 0 } = absolute \ permittivity = 8.854 \times { 10 }^{ -12 }F/m $$ $$ { \epsilon }_{ r } = 1 \qquad \qquad \left[ \because medium \ is \ air \right] $$ d = distance between the charges = 1m

The magnitude of force between two charges is given by $$ \boxed { F = \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } } $$ Lets Substitute the values, $$ \therefore F = \frac { 1 }{ 4 \times 3.14 \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 } \frac { 1 \times 1 }{ { \left( 1 \right) }^{ 2 } } $$ $$ F = \frac { 1 }{ 4 \times 3.14 \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 } $$ $$ F = \frac { 1 }{ 0.111 \times { 10 }^{ -9 } } $$ $$ F = 8.992 \times { 10 }^{ 9 } N $$

An electron and a proton are at a distance of 10^-9 m from each other in a free space. Compute the force between them.

Solution:We have, The charge on an electron, \({ q }_{ 1 } = -1.6 \times { 10 }^{ -19 } \) C
and the charge on a proton, \({ q }_{ 2 } = +1.6 \times { 10 }^{ -19 }\) C. $$ { q }_{ 1 } = { q }_{ 2 } = 1 \ coulomb $$ $$ { \epsilon }_{ 0 } = absolute \ permittivity = 8.854 \times { 10 }^{ -12 }F/m $$ $$ { \epsilon }_{ r } = 1 \qquad \qquad \left[ \because medium \ is \ air \right] $$ d = distance between the charges = \({ 10 }^{ -9 }\) m

The magnitude of force between two charges is given by $$ \boxed { F = \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } } $$ Lets Substitute the values, $$ \therefore F = \left( \frac { 1 }{ 4 \times 3.14 \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 } \right) \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } $$ $$ \therefore F = \left( \frac { 1 }{ 0.111 \times { 10 }^{ -9 } } \right) \left( \frac { \left( -1.6 \times { 10 }^{ -19 } \right) \times \left( +1.6 \times { 10 }^{ -19 } \right) }{ { \left( { 10 }^{ -9 } \right) }^{ 2 } } \right) $$ $$ \therefore F = \frac { 1 }{ 0.111 \times { 10 }^{ -9 } } \left( -2.56 \times { 10 }^{ -20 } \right) $$ $$ \therefore F = -23.063 \times { 10 }^{ -11 } N $$ The negative sign indicates that the force is attractive in nature.

Coulomb performed a number of experiments to see the effect of placing two small charges near each other. On his experimental researches, he established a law which is known as Coulombs Law of Electrostatics.

According to Coulombs Law :The force of attraction or repulsion between two charges is directly proportional to the product of magnitude of the two charges and inversely proportional to the square of distance between them.

If two point charges \({ q }_{ 1 } \) and \({ q }_{ 2 }\) have distance d between them, then force F between the charges can be mathematically expressed as: $$ F \propto { q }_{ 1 }{ q }_{ 2 }\qquad ...(1)$$ $$ F \propto \frac { 1 }{ { d }^{ 2 } } \qquad ...(2)$$ Combining (1) and (2), $$ F \propto \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } or F = K\frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } \qquad ...(3) $$ Where, \(K\) is a constant of proportionality and its value depends upon the medium in which the charges are placed and the system of units used.

In SI units force is measured in newton, charge in coulomb, distance in metre and the value of \(K\) is given as, $$ K = \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \qquad ...(4) (in SI system)$$ where, \({ \epsilon }_{ 0 }\) = Absolute permittivity of vacuum or permittivity of free space = \(8.854 \times { 10 }^{ -12 }\) farad/metre
\( { \epsilon }_{ r }\) = Relative permittivity of the medium w.r.t. vacuum in which the charges are placed (e.g. for air \({ \epsilon }_{ r }\) = 1)

Putting the value of \(K\) in equation (3), we get, $$ \boxed { F = \left( \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \right) \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } } \qquad ... (5)$$ Unit Charge or one coulomb charge in SI system can be defined as the amount of charge which when placed at a distance of one metre from an equal and similar charge in air, is repelled with a force of \(9 \times { 10 }^{ 9 } \)newton from it.

As per this defination, $$ { q }_{ 1 } = { q }_{ 2 } = q, $$ $$F = 9 \times { 10 }^{ 9 } newton,$$ $$ { \epsilon }_{ 0 } = 8.854 \times { 10 }^{ -12 } farad/metre,$$ \({ \epsilon }_{ r } = 1\) (Assuming two charges to be placed in air)
d = 1m

Putting these values in the above relation, we get $$ 9 \times { 10 }^{ 9 } = \frac { q \times q }{ 4\pi \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 \times { 1 }^{ 2 } } $$ $$ 9 \times { 10 }^{ 9 } = \frac { { q }^{ 2 } }{ 4\pi \times \left( 8.854 \times { 10 }^{ -12 } \right) } $$ $$ { q }^{ 2 } = \left( 9 \times { 10 }^{ 9 } \right) \times 4\pi \times \left( 8.854 \times { 10 }^{ -12 } \right) = 1.00086 $$ $$ \therefore q = \pm 1.00043 \ coulomb $$ Hence, unit charge.

Determine the force between two charges, each of one coulomb when they are separated at one metre distance in air.

Solution:We have, $${ q }_{ 1 } = { q }_{ 2 } = 1 \ coulomb $$ $$ { \epsilon }_{ 0 } = absolute \ permittivity = 8.854 \times { 10 }^{ -12 }F/m $$ $$ { \epsilon }_{ r } = 1 \qquad \qquad \left[ \because medium \ is \ air \right] $$ d = distance between the charges = 1m

The magnitude of force between two charges is given by $$ \boxed { F = \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } } $$ Lets Substitute the values, $$ \therefore F = \frac { 1 }{ 4 \times 3.14 \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 } \frac { 1 \times 1 }{ { \left( 1 \right) }^{ 2 } } $$ $$ F = \frac { 1 }{ 4 \times 3.14 \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 } $$ $$ F = \frac { 1 }{ 0.111 \times { 10 }^{ -9 } } $$ $$ F = 8.992 \times { 10 }^{ 9 } N $$

An electron and a proton are at a distance of 10^-9 m from each other in a free space. Compute the force between them.

Solution:We have, The charge on an electron, \({ q }_{ 1 } = -1.6 \times { 10 }^{ -19 } \) C
and the charge on a proton, \({ q }_{ 2 } = +1.6 \times { 10 }^{ -19 }\) C. $$ { q }_{ 1 } = { q }_{ 2 } = 1 \ coulomb $$ $$ { \epsilon }_{ 0 } = absolute \ permittivity = 8.854 \times { 10 }^{ -12 }F/m $$ $$ { \epsilon }_{ r } = 1 \qquad \qquad \left[ \because medium \ is \ air \right] $$ d = distance between the charges = \({ 10 }^{ -9 }\) m

The magnitude of force between two charges is given by $$ \boxed { F = \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } } $$ Lets Substitute the values, $$ \therefore F = \left( \frac { 1 }{ 4 \times 3.14 \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 } \right) \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } $$ $$ \therefore F = \left( \frac { 1 }{ 0.111 \times { 10 }^{ -9 } } \right) \left( \frac { \left( -1.6 \times { 10 }^{ -19 } \right) \times \left( +1.6 \times { 10 }^{ -19 } \right) }{ { \left( { 10 }^{ -9 } \right) }^{ 2 } } \right) $$ $$ \therefore F = \frac { 1 }{ 0.111 \times { 10 }^{ -9 } } \left( -2.56 \times { 10 }^{ -20 } \right) $$ $$ \therefore F = -23.063 \times { 10 }^{ -11 } N $$ The negative sign indicates that the force is attractive in nature.

The number of electrons in an atom is equal to the number of protons, therefore, atom is neutral as a whole. A body consists of atoms, therefore, the body is neutral under ordinary conditions. However, if from such a neutral body, electrons are removed, there occurs a shortage of electrons in the body.

Consequently the body no longer remains neutral. The result is that the body attains positive charge.

Thus, when a body is having shortage of electrons, it is said to be positively charged.

On the other hand, a negatively charged body has excess of electrons from its normal due share.

Total deficiency or excess of electrons in a body is known as charge on the body.

To give a negative charge to any body, extra electrons must be supplied to it.

To supply these extra electrons, work will have to be done, which is stored in the body in the form of energy.

This makes the charged body capable of doing work.

The charge on an electron is so small that it is not possible and convenient to take it as the unit. In practice, the charge is measured in coulomb (C).

1 coulomb of charge = The charge on \(625 \times {10}^{16}\) electrons.

Therefore, practical unit of charge is coulomb.

Like charges repel each other while unlike charges attract each other, means when two like charges are brought near to each other, they will repel each other and if two unlike charges [a +ve and a -ve charge] are brought near to each other, they will start attracting each other.

The number of electrons in an atom is equal to the number of protons, therefore, atom is neutral as a whole. A body consists of atoms, therefore, the body is neutral under ordinary conditions. However, if from such a neutral body, electrons are removed, there occurs a shortage of electrons in the body.

Consequently the body no longer remains neutral. The result is that the body attains positive charge.

Thus, when a body is having shortage of electrons, it is said to be positively charged.

On the other hand, a negatively charged body has excess of electrons from its normal due share.

Total deficiency or excess of electrons in a body is known as charge on the body.

To give a negative charge to any body, extra electrons must be supplied to it.

To supply these extra electrons, work will have to be done, which is stored in the body in the form of energy.

This makes the charged body capable of doing work.

The charge on an electron is so small that it is not possible and convenient to take it as the unit. In practice, the charge is measured in coulomb (C).

1 coulomb of charge = The charge on \(625 \times {10}^{16}\) electrons.

Therefore, practical unit of charge is coulomb.

Like charges repel each other while unlike charges attract each other, means when two like charges are brought near to each other, they will repel each other and if two unlike charges [a +ve and a -ve charge] are brought near to each other, they will start attracting each other.

The method was discovered by Craig and Post. It is consists of 300-400 such chambers. The organic solvent and the aqueous solution is introduced to tube A and then it passes to B. It is shaken and is allowed to attain the equilibrium. Now, the apparatus is tilted so that the upper layer gets decanted through C and is collected in D. When the apparatus is again made vertical. The liquid passes through D into E in the next chamber of A and then to B. The process is repeated till the two liquids gets almost separated.

Application:

1. With the help of this method we can have accurate quantitative analysis of a single as well as the mixture of the components.

2. In this case apparatus required are very simple (separating funnel burette pipets conical flask etc.)

3. Time required for analysis is very small.

4. The method is very well used for detection of traces quantity of substance where precipitation method (Gravimetry) is not possible.

5. Fe⁺³ ferric ion can be easily extracted by ether from 6 molar HCl solution of the ferrous ally and iron ore.

6. The extraction can also be use in the extraction of metal as metal chelate where later has high solubility in an immiscible solvent such as chloroform and benzene.

7. In industrial and commercial field extraction by counter current extraction is frequently applied in the separation of components where the difference in the distribution coefficient are small.

8. The phenomenon is widely applied in drug analysis.

9. The solvent extraction is used in clinical laboratory.

10. Metal chelates are more soluble in non-polar solvents. Thus Ni(II) in its tetra co-ordinate complex with dimethyl glyoxime can be extracted into chloroform. In presence of citrate or tartrate the precipitation of Fe(III) and Cr(III) can be avoided.

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The method was discovered by Craig and Post. It is consists of 300-400 such chambers. The organic solvent and the aqueous solution is introduced to tube A and then it passes to B. It is shaken and is allowed to attain the equilibrium. Now, the apparatus is tilted so that the upper layer gets decanted through C and is collected in D. When the apparatus is again made vertical. The liquid passes through D into E in the next chamber of A and then to B. The process is repeated till the two liquids gets almost separated.

Application:

1. With the help of this method we can have accurate quantitative analysis of a single as well as the mixture of the components.

2. In this case apparatus required are very simple (separating funnel burette pipets conical flask etc.)

3. Time required for analysis is very small.

4. The method is very well used for detection of traces quantity of substance where precipitation method (Gravimetry) is not possible.

5. Fe⁺³ ferric ion can be easily extracted by ether from 6 molar HCl solution of the ferrous ally and iron ore.

6. The extraction can also be use in the extraction of metal as metal chelate where later has high solubility in an immiscible solvent such as chloroform and benzene.

7. In industrial and commercial field extraction by counter current extraction is frequently applied in the separation of components where the difference in the distribution coefficient are small.

8. The phenomenon is widely applied in drug analysis.

9. The solvent extraction is used in clinical laboratory.

10. Metal chelates are more soluble in non-polar solvents. Thus Ni(II) in its tetra co-ordinate complex with dimethyl glyoxime can be extracted into chloroform. In presence of citrate or tartrate the precipitation of Fe(III) and Cr(III) can be avoided.

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Though the multiple extraction is more beneficial than single step extraction. For analytical chemist it is easy to find out the amount extracted in single extraction, so he follow single extraction rather than multiple extraction because it is tedious. The relation between percentage extraction (E) for single extraction can be derived as follow: We know $$ { W }_{ 1 } = \left[ \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right] \times a \qquad ...(1) $$ where \({ W }_{ 1 }\) is the weight of solute remaining after 1st extraction.
\( { V }_{ W } \)= Volume of Aqueous Phase
\( { V }_{ o } \) = Volume of Organic Phase
\(D\) = Distribution Ratio
\(a\) = Total weight of Solute present initially.

The amount extracted will be a $$ { W }_{ 1 } = a - a\left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) \qquad ...(2) $$ $$ = a - a\left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) $$ $$ \therefore Percentage \ extraction \ (E) = \left( \frac { a - { W }_{ 1 } }{ a } \right) \times 100 $$ Lets Substitute the values from equation (1) $$ E = \left( \frac { a - a\left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) }{ a } \right) \times 100 $$ $$ E = \left( 1 - \left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) \right) \times 100 $$ $$ E = \left( \frac { { V }_{ o }D + { V }_{ W } - { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) \times 100 $$ $$E = \left( \frac { { V }_{ o }D }{ { V }_{ o }D + { V }_{ W } } \right) \times 100 $$ on dividing the numerator and denominator of RHS by \({ V }_{ o }\) we get $$ E = \left( \frac { D }{ D + \frac { { V }_{ W } }{ { V }_{ o } } } \right) \times 100 $$ $$ \therefore Percentage extraction (E) = \frac { 100D }{ D + \frac { { V }_{ W } }{ { V }_{ o } } } $$ Thus % extraction (E) depends on D as well as \(\frac { { V }_{ W } }{ { V }_{ o } } \).

Though the multiple extraction is more beneficial than single step extraction. For analytical chemist it is easy to find out the amount extracted in single extraction, so he follow single extraction rather than multiple extraction because it is tedious. The relation between percentage extraction (E) for single extraction can be derived as follow: We know $$ { W }_{ 1 } = \left[ \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right] \times a \qquad ...(1) $$ where \({ W }_{ 1 }\) is the weight of solute remaining after 1st extraction.
\( { V }_{ W } \)= Volume of Aqueous Phase
\( { V }_{ o } \) = Volume of Organic Phase
\(D\) = Distribution Ratio
\(a\) = Total weight of Solute present initially.

The amount extracted will be a $$ { W }_{ 1 } = a - a\left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) \qquad ...(2) $$ $$ = a - a\left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) $$ $$ \therefore Percentage \ extraction \ (E) = \left( \frac { a - { W }_{ 1 } }{ a } \right) \times 100 $$ Lets Substitute the values from equation (1) $$ E = \left( \frac { a - a\left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) }{ a } \right) \times 100 $$ $$ E = \left( 1 - \left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) \right) \times 100 $$ $$ E = \left( \frac { { V }_{ o }D + { V }_{ W } - { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) \times 100 $$ $$E = \left( \frac { { V }_{ o }D }{ { V }_{ o }D + { V }_{ W } } \right) \times 100 $$ on dividing the numerator and denominator of RHS by \({ V }_{ o }\) we get $$ E = \left( \frac { D }{ D + \frac { { V }_{ W } }{ { V }_{ o } } } \right) \times 100 $$ $$ \therefore Percentage extraction (E) = \frac { 100D }{ D + \frac { { V }_{ W } }{ { V }_{ o } } } $$ Thus % extraction (E) depends on D as well as \(\frac { { V }_{ W } }{ { V }_{ o } } \).

If a solution contains two or more solutes say A and B, it is observed that when A is extracted, some amount of B is also extracted. The extent of the seperation can be expressed in terms of one factor called Seperation Factor \(\beta\) . This is related to the distribution ratio of A and B. $$ \boxed { \beta = \frac { D_{ A } }{ { D }_{ B } } = \frac { { { C }_{ o(A) } }/{ { C }_{ a(A) } } }{ { { C }_{ o(B) } }/{ { C }_{ a(B) } } } } $$ It is ratio therefore no unit and no dimension.

The larger value is always placed in the numerator. \beta must be made as large as possible by choice of extractant and by adjusting the volume ratio.

When \(D_{ A }\) = 10 and \(D_{ B } \)= 0.1. The \(\beta = \frac { 10 }{ 0.1 } = 100%\)

single extraction in case will remove 91% of A and 9% of B. It can be obtained for A, $$ E = \left[ \frac { 100{ D }_{ A } }{ { D }_{ A } + { { V }_{ W } }/{ { V }_{ o } } } \right] $$ for $${ V }_{ W } = { V }_{ o }\qquad \qquad \therefore { { V }_{ W } }/{ { V }_{ o } } = 1 $$ $$ \therefore E = \left[ \frac { 100 \times 10 }{ 10 + 1 } \right] = \left[ \frac { 1000 }{ 11 } \right] = 90.9% $$ $$ Similarly \ for \ B, \ E = \left[ \frac { 100{ D }_{ B } }{ { D }_{ B } + { { V }_{ W } }/{ { V }_{ o } } } \right] $$ $$ for \ { V }_{ W } = { V }_{ o }\qquad \qquad \therefore { { V }_{ W } }/{ { V }_{ o } } = 1 $$ $$ \therefore E = \left[ \frac { 100 \times 0.1 }{ 0.1 + 1 } \right] = \left[ \frac { 10 }{ 1.1 } \right] = 9.1% $$ The seperation of A is almost complete from B if the seperation factor B is high. It can be only in the case when \({ D }_{ A }\) is large and \({ D }_{ B }\) is small. For a given value of \({ D }_{ A }\) and \({ D }_{ B }\), the seperation effect can be increased by adjusting the volume ratio given by Nush Densen Equation which says. $$ \boxed { \frac { { V }_{ o } }{ { V }_{ W } } = \frac { 1 }{ { \left( { D }_{ A }{ D }_{ B } \right) }^{ { 1 }/{ 2 } } } } $$

If a solution contains two or more solutes say A and B, it is observed that when A is extracted, some amount of B is also extracted. The extent of the seperation can be expressed in terms of one factor called Seperation Factor \(\beta\) . This is related to the distribution ratio of A and B. $$ \boxed { \beta = \frac { D_{ A } }{ { D }_{ B } } = \frac { { { C }_{ o(A) } }/{ { C }_{ a(A) } } }{ { { C }_{ o(B) } }/{ { C }_{ a(B) } } } } $$ It is ratio therefore no unit and no dimension.

The larger value is always placed in the numerator. \beta must be made as large as possible by choice of extractant and by adjusting the volume ratio.

When \(D_{ A }\) = 10 and \(D_{ B } \)= 0.1. The \(\beta = \frac { 10 }{ 0.1 } = 100%\)

single extraction in case will remove 91% of A and 9% of B. It can be obtained for A, $$ E = \left[ \frac { 100{ D }_{ A } }{ { D }_{ A } + { { V }_{ W } }/{ { V }_{ o } } } \right] $$ for $${ V }_{ W } = { V }_{ o }\qquad \qquad \therefore { { V }_{ W } }/{ { V }_{ o } } = 1 $$ $$ \therefore E = \left[ \frac { 100 \times 10 }{ 10 + 1 } \right] = \left[ \frac { 1000 }{ 11 } \right] = 90.9% $$ $$ Similarly \ for \ B, \ E = \left[ \frac { 100{ D }_{ B } }{ { D }_{ B } + { { V }_{ W } }/{ { V }_{ o } } } \right] $$ $$ for \ { V }_{ W } = { V }_{ o }\qquad \qquad \therefore { { V }_{ W } }/{ { V }_{ o } } = 1 $$ $$ \therefore E = \left[ \frac { 100 \times 0.1 }{ 0.1 + 1 } \right] = \left[ \frac { 10 }{ 1.1 } \right] = 9.1% $$ The seperation of A is almost complete from B if the seperation factor B is high. It can be only in the case when \({ D }_{ A }\) is large and \({ D }_{ B }\) is small. For a given value of \({ D }_{ A }\) and \({ D }_{ B }\), the seperation effect can be increased by adjusting the volume ratio given by Nush Densen Equation which says. $$ \boxed { \frac { { V }_{ o } }{ { V }_{ W } } = \frac { 1 }{ { \left( { D }_{ A }{ D }_{ B } \right) }^{ { 1 }/{ 2 } } } } $$

In choosing a proper solvent for solvent extractions, the following factors must be taken into consideration.

1. The solvent selected should be such that, the solubility of the solute to be extracted is more. This means seperation factor (B) must be large.

2. The two phases used must be totally immiscible.

3. The solvent must not react chemically with the solute present in aqueous solution.

4. The solvent must be recoverable without much expenditure.

5. The density of the solvent must be different than aqueous phase it will help in quick settling of the two phases.

6. The solvent should not be toxic.

7. The solvent should be cheap.

8. The solvent should possess low inflammability.

9. The solvent should have low viscosity, low vapour pressure and low freezing point.

10. The solvent should be readily available.

In choosing a proper solvent for solvent extractions, the following factors must be taken into consideration.

1. The solvent selected should be such that, the solubility of the solute to be extracted is more. This means seperation factor (B) must be large.

2. The two phases used must be totally immiscible.

3. The solvent must not react chemically with the solute present in aqueous solution.

4. The solvent must be recoverable without much expenditure.

5. The density of the solvent must be different than aqueous phase it will help in quick settling of the two phases.

6. The solvent should not be toxic.

7. The solvent should be cheap.

8. The solvent should possess low inflammability.

9. The solvent should have low viscosity, low vapour pressure and low freezing point.

10. The solvent should be readily available.

In general, organic solute (covalent solute) are more soluble in non polar (organic) solvents than water and hydrated salts are more soluble in water (polar) solvent. The inorganic salts in the solution are in the form of ions. In solvent extraction, we must see that solubility of inorganic salts in water should be less and solubility in organic solvent must be increased. The solubility of the metal ions in water is due to the association of water molecules with charged metal ions. in other words due to the hydration. Therefore, if the charged on the metal ions is neutralised, the solubility of the metal ions in the watt will be decreased. The charge on the metal ion is neutralised by reaction of metal ions with the appropriate complexing agent to form the metal-chellate complex. The bulkier the complex and if it is more hydrophobic in nature better will be extraction. .. Each ion will have to be treated differently. Thus, aluminium ions in the solution are usually converted into hydroxy aluminium quinolate by the addition of 8-hydroxy quinolene. The chellate thus formed can be extracted by using benzene or chloroform. It is possible to extract aluminium at pH 5:18, 3 in the presence of iron Nickel and Vanadium. The other complexing agents are dimethyl - glyoxime, acetyl acetone cupferron, \(\alpha\) - nitroso \(\beta\) - naphthol dithizone etc.

In general, organic solute (covalent solute) are more soluble in non polar (organic) solvents than water and hydrated salts are more soluble in water (polar) solvent. The inorganic salts in the solution are in the form of ions. In solvent extraction, we must see that solubility of inorganic salts in water should be less and solubility in organic solvent must be increased. The solubility of the metal ions in water is due to the association of water molecules with charged metal ions. in other words due to the hydration. Therefore, if the charged on the metal ions is neutralised, the solubility of the metal ions in the watt will be decreased. The charge on the metal ion is neutralised by reaction of metal ions with the appropriate complexing agent to form the metal-chellate complex. The bulkier the complex and if it is more hydrophobic in nature better will be extraction. .. Each ion will have to be treated differently. Thus, aluminium ions in the solution are usually converted into hydroxy aluminium quinolate by the addition of 8-hydroxy quinolene. The chellate thus formed can be extracted by using benzene or chloroform. It is possible to extract aluminium at pH 5:18, 3 in the presence of iron Nickel and Vanadium. The other complexing agents are dimethyl - glyoxime, acetyl acetone cupferron, \(\alpha\) - nitroso \(\beta\) - naphthol dithizone etc.

It is the simplest and most widely used method. The method is to be applied when the distribution D is high.

Requirement: Simple glass separating flask beaker, funnel etc.

Experiment: In this technique the aqueous phase containing the solute to be extracted is brought in contact with the immiscible organic solvent in a simple separating funnel. The two phases are shaken vigorously at least for 10-15 minutes. Then are allowed to attain the equilibrium. After the layers become clear, the denser liquid is withdrawn by opening the stop cock of the funnel. Once the denser liquid is separated, then lighter layer is separated.

It is the simplest and most widely used method. The method is to be applied when the distribution D is high.

Requirement: Simple glass separating flask beaker, funnel etc.

Experiment: In this technique the aqueous phase containing the solute to be extracted is brought in contact with the immiscible organic solvent in a simple separating funnel. The two phases are shaken vigorously at least for 10-15 minutes. Then are allowed to attain the equilibrium. After the layers become clear, the denser liquid is withdrawn by opening the stop cock of the funnel. Once the denser liquid is separated, then lighter layer is separated.

When the distribution is small, we have to carry out the multiple extractions. But multiple extraction brings no. of the complications and is very tedious. Kutcher and Steudel devised a continuous extractions to avoid the multiple extractions.

1. When the extracting solvent is heavier than water (aqueous). It is consist of tube and reservoir. The reservoir contain the same organic solvent. The tube is fitted with the condenser. The solvent in reservoir is heated slowly which is converted into vapours, Which move up and are condensed and the organic solvent is converted into liquid which is heavy and descend while moving through aqueous, the solute gets transferred from aqueous to organic and at the bottom we get solution of solute in organic. so the level increases and corresponding solution will fall in reservoir. This continue till maximum amount of solute is transferred from aqueous to organic. Then by opening the stopper at the bottom, we can separate organic and aqueous layer, then solute can be extracted from organic.

2. When organic solvent is lighter than aqueous. In that case reservoir contain organic, which on warming is converted into to the vapours, which ascends and is condensed in condenser which descends through bubbler type of funnel. This liquids goes to bottom but as it is light so it ascends, and passes through solution and here the solute gets transferred from aqueous solution. So the level of of the solution increases and the corresponding gets transferred to reservoir.

When the distribution is small, we have to carry out the multiple extractions. But multiple extraction brings no. of the complications and is very tedious. Kutcher and Steudel devised a continuous extractions to avoid the multiple extractions.

1. When the extracting solvent is heavier than water (aqueous). It is consist of tube and reservoir. The reservoir contain the same organic solvent. The tube is fitted with the condenser. The solvent in reservoir is heated slowly which is converted into vapours, Which move up and are condensed and the organic solvent is converted into liquid which is heavy and descend while moving through aqueous, the solute gets transferred from aqueous to organic and at the bottom we get solution of solute in organic. so the level increases and corresponding solution will fall in reservoir. This continue till maximum amount of solute is transferred from aqueous to organic. Then by opening the stopper at the bottom, we can separate organic and aqueous layer, then solute can be extracted from organic.

2. When organic solvent is lighter than aqueous. In that case reservoir contain organic, which on warming is converted into to the vapours, which ascends and is condensed in condenser which descends through bubbler type of funnel. This liquids goes to bottom but as it is light so it ascends, and passes through solution and here the solute gets transferred from aqueous solution. So the level of of the solution increases and the corresponding gets transferred to reservoir.