Refraction of light through a thin lens : Lens maker's formula

Derivation of refraction of light through a thin lens & Lens maker's formula:

Let us consider, A convex lens having thickness $t$ and radius of curvature of surfaces is $R_{1}$ and $R_{2}$. If an object $O$ is placed at distance $u$ from the first surface of the convex lens and its image $I'$ is formed at distance $v'$ from the first surface of the convex lens then refraction of light through the first spherical surface of the lens

$ \frac{\left( n_{2} - n_{1} \right)}{R_{1}} = \frac{n_{2}}{v'} - \frac{n_{1}}{u} \qquad(1) $

Now the Image $I'$ works as a virtual object for the second surface of the convex lens which image $I$ formed at distance $v$ from the second surface of the lens. So refraction of light through the second surface of the lens

$ \frac{\left( n_{1} - n_{2} \right)}{R_{2}} = \frac{n_{1}}{v} - \frac{n_{2}}{v' - t} $

Here $t$ is the thickness of the lens. If the lens is very thin then thickness will be $t=0$. Therefore above equation for second surface of the lens can be written as

$ \frac{\left( n_{1} - n_{2} \right)}{R_{2}} = \frac{n_{1}}{v} - \frac{n_{2}}{v'} \qquad(2) $

Now add the equation $(1)$ and equation $(2)$, So

$\frac{\left( n_{2} - n_{1} \right)}{R_{1}} + \frac{\left( n_{1} - n_{2} \right)}{R_{2}} = \frac{n_{2}}{v'} - \frac{n_{1}}{u} + \frac{n_{1}}{v} - \frac{n_{2}}{v'} $

$\frac{\left( n_{2} - n_{1} \right)}{R_{1}} + \frac{\left( n_{1} - n_{2} \right)}{R_{2}} = - \frac{n_{1}}{u} + \frac{n_{1}}{v} $

$\left( n_{2} - n_{1} \right) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right) =n_{1} \left( \frac{1}{v} - \frac{1}{u} \right) $

$n_{1} \left( \frac{1}{v} - \frac{1}{u} \right) = \left( n_{2} - n_{1} \right) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right) $

$ \left( \frac{1}{v} - \frac{1}{u} \right) = \frac{\left( n_{2} - n_{1} \right)}{n_{1}} \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right) $

We know that the equation of the focal length of a lens

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

Where $f \rightarrow$ Focal length of convex lens. Now substitute the value of $f$ in the above equation

$ \frac{1}{f} = \frac{\left( n_{2} - n_{1} \right)}{n_{1}} \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$

$ \frac{1}{f} = \left( \frac{n_{2}}{n_{1}} - 1 \right) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$

$ \frac{1}{f} = \left( n - 1 \right) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right) \qquad \left( \because \frac{n_{2}}{n_{1}}= n\right)$

The above equation represents the equation of refraction of light through a thin lens and lens maker's formula.

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Refraction of light through the convex spherical surface

Derivation of refraction of light through the convex spherical surface:

Let us consider, a convex spherical surface which has radius of curvature $R$. If an object $O$ is placed at a distance $u$ from pole $P$ and its image $I$ is formed at distance $v$ from pole $P$ and the angle subtended by the object, image, and center of curvature is $\alpha$, $\beta$, and $\gamma$ then from figure

In $\Delta MOC$

$i= \gamma + \alpha \qquad(1)$

In $\Delta MIC$

$r = \gamma + \beta \qquad(2)$

According to Snell's Law:

$\frac{sin \: i}{sin \: r} = \frac{n_{2}}{n_{1}} \qquad(3)$

Here the aperture of the spherical surface is very small so point $M$ will be very close to point $P$ and angle $i$ and $r$ will be small. So

$sin \: i \approx i$
$sin \: r \approx r$

So equation $(3)$ can be written as

$\frac{ i}{ r} = \frac{n_{2}}{n_{1}} \qquad(4)$

Now subtitute the value of $i$ and $r$ from equation $(1)$ and equation $(2)$ in equation $(4)$

$\frac{ \left( \gamma + \alpha \right)}{\left( \gamma + \beta \right)} = \frac{n_{2}}{n_{1}} $

$ \left( \gamma + \alpha \right) n_{1} = \left( \gamma + \beta \right) n_{2} $

$ \gamma \: n_{1} + \alpha \: n_{1} = \gamma \: n_{2} + \beta \: n_{2} $

$ \gamma \: n_{1} - \gamma \: n_{2} = \beta \: n_{2} - \alpha \: n_{1} $

$ \left( n_{1} - n_{2} \right) \gamma = \beta \: n_{2} - \alpha \: n_{1} \qquad(5)$

Again here the aperture of the spherical surface is very small so point $N$ will be very close to point $P$. Therefore the angle $\alpha$, $\beta$, and $\gamma$ will be very small. Then in $\Delta MON$

$\alpha = tan \alpha = \frac {MN}{ON} \approx \frac{MN}{OP} = \frac{h}{-u}$

Similarly for angle $\beta$ and $\gamma$, Apply in $\Delta MIN$ and $\Delta MCN$

$\beta = tan \beta = \frac {MN}{IN} \approx \frac{MN}{IP} = \frac{h}{-v}$

$\gamma = tan \gamma = \frac {MN}{CN} \approx \frac{MN}{CP} = \frac{h}{R}$

Now subtitute the value of $\alpha$, $\beta$, and $\gamma$ from above in equation $(5)$

$ \left( \frac{h}{R} \right) \left( n_{1} - n_{2} \right) = \left( \frac{h}{-v} \right) n_{2} - \left( \frac{h}{-u} \right) n_{1} $

$ \frac{\left( n_{2} - n_{1} \right)}{R} = \frac{n_{2}}{v} - \frac{n_{1}}{u} $

This is the equation of refraction of light through a convex spherical surface.

Another form of the above equation:

$ n_{1} \frac{\left( \frac{n_{2}}{n_{1}} - 1 \right)}{R} = n_{1} \frac{\frac{n_{2}}{n_{1}}}{v} - \frac{1}{u} $

$ n_{1} \frac{\left( \frac{n_{2}}{n_{1}} - 1 \right)}{R} = n_{1} \frac{\left(\frac{n_{2}}{n_{1}}\right)}{v} - \frac{1}{u} $

$ \frac{\left( \frac{n_{2}}{n_{1}} - 1 \right)}{R} = \frac{\left(\frac{n_{2}}{n_{1}}\right)}{v} - \frac{1}{u} $

$ \frac{\left( n - 1 \right)}{R} = \frac{n}{v} - \frac{1}{u} \qquad \left( \because \frac{n_{2}}{n_{1}}= n\right) $

Where $n \rightarrow Refractive \: index \: of \: glass \: with \: respect \: to \: air$

This is another form of the equation of refraction of light through a convex spherical surface.

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Refraction of light through the concave spherical surface

Derivation of refraction of light through the concave spherical surface:

Let us consider, a concave spherical surface of radius of curvature $R$. If an object $O$ is placed at a distance $u$ from pole $P$ and its image $I$ is formed at distance $v$ from pole $P$ and the angle subtended by the object, image, and center of curvature is $\alpha$, $\beta$, and $\gamma$ then from figure

In $\Delta MOC$

$\gamma= \alpha + i $

$i= \gamma - \alpha \qquad(1)$

In $\Delta MIC$

$\gamma= \beta + r $

$r = \gamma - \beta \qquad(2)$

According to Snell's Law:

$\frac{sin \: i}{sin \: r} = \frac{n_{2}}{n_{1}} \qquad(3)$

Here the aperture of the spherical surface is very small so point $M$ will be very close to point $P$ and angle $i$ and $r$ will be small. So

$sin \: i \approx i$
$sin \: r \approx r$

So equation $(3)$ can be written as

$\frac{ i}{ r} = \frac{n_{2}}{n_{1}} \qquad(4)$

Now subtitute the value of $i$ and $r$ from equation $(1)$ and equation $(2)$ in equation $(4)$

$\frac{ \left( \gamma - \alpha \right)}{\left( \gamma - \beta \right)} = \frac{n_{2}}{n_{1}} $

$ \left( \gamma - \alpha \right) n_{1} = \left( \gamma - \beta \right) n_{2} $

$ \gamma \: n_{1} - \alpha \: n_{1} = \gamma \: n_{2} - \beta \: n_{2} $

$ \gamma \: n_{1} - \gamma \: n_{2} = -\beta \: n_{2} + \alpha \: n_{1} $

$ \left( n_{1} - n_{2} \right) \gamma = -\beta \: n_{2} + \alpha \: n_{1} \qquad(5)$

Again here the aperture of the spherical surface is very small so point $N$ will be very close to point $P$. Therefore the angle $\alpha$, $\beta$, and $\gamma$ will be very small. Then in $\Delta MON$

$\alpha = tan \alpha = \frac {MN}{ON} \approx \frac{MN}{OP} = \frac{h}{-u}$

Similarly for angle $\beta$ and $\gamma$, Apply in $\Delta MIN$ and $\Delta MCN$

$\beta = tan \beta = \frac {MN}{IN} \approx \frac{MN}{IP} = \frac{h}{-v}$

$\gamma = tan \gamma = \frac {MN}{CN} \approx \frac{MN}{CP} = \frac{h}{-R}$

Now subtitute the value of $\alpha$, $\beta$, and $\gamma$ from above in equation $(5)$

$ \left( \frac{h}{-R} \right) \left( n_{1} - n_{2} \right) = - \left( \frac{h}{-v} \right) n_{2} + \left( \frac{h}{-u} \right) n_{1} $

$ \frac{\left( n_{2} - n_{1} \right)}{R} = \frac{n_{2}}{v} - \frac{n_{1}}{u} $

This is the equation of refraction of light through a concave spherical surface.

Another form of the above equation:

$ n_{1} \frac{\left( \frac{n_{2}}{n_{1}} - 1 \right)}{R} = n_{1} \frac{\frac{n_{2}}{n_{1}}}{v} - \frac{1}{u} $

$ n_{1} \frac{\left( \frac{n_{2}}{n_{1}} - 1 \right)}{R} = n_{1} \frac{\left(\frac{n_{2}}{n_{1}}\right)}{v} - \frac{1}{u} $

$ \frac{\left( \frac{n_{2}}{n_{1}} - 1 \right)}{R} = \frac{\left(\frac{n_{2}}{n_{1}}\right)}{v} - \frac{1}{u} $

$ \frac{\left( n - 1 \right)}{R} = \frac{n}{v} - \frac{1}{u} \qquad \left( \because \frac{n_{2}}{n_{1}}= n\right) $

Where $n \rightarrow Refractive \: index \: of \: glass \: with \\ respect \: to \: air$

This is another form of the equation of refraction of light through a concave spherical surface.

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Physical Significance of Maxwell's Equations

Physical Significance:

The physical significance of Maxwell's equations obtained from integral form are given below:

Maxwell's First Equation:

1. The total electric displacement through the surface enclosing a volume is equal to the total charge within the volume.

2. It represents Gauss Law.

3. This law is independent of time. Charge acts as source or sink for the lines of electric force.

Maxwell's Second Equation:

1. The total magnetic flux emitting through any closed surface is zero. An isolated magnet do not exist monopoles.

2. There is no source or sink for lines of magnetic force.

3. This is time independent equation.

Maxwell's Third Equation:

1. The electromotive force around the closed path is equal to the time derivative of the magnetic displacement through any surface bounded by the path.

2. This gives relation between electric field $E$ and magnetic induction $B$.

3. This expression is time varying i.e. $E$ is generated by time variation of $B$.

4.This gives relation in variation of E with the time variation of $B$ or $H$.

5. This is a mathematical form of Faraday's law of electromagnetic induction or Lenz's Law.

Maxwell's Fourth Equation:

1. The magneto-motive force around the closed path is equal to the conduction current plus displacement current through any surface bounded by the path.

2. This is time dependent wave equation.

3 This is a mathematical form of Ampere circuital law.

4. Magnetic induction $B$ can be generated from $J$ and time variation of magnetic displacement $D$.

5. This relates the space variation of $B$ with time variation of $D$.

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Circuit containing Inductor and Capacitor in Series (L-C Series Circuit )

Mathematical Analysis of L-C Series Circuit :

Let us consider, a circuit containing inductor $L$ capacitor $C$ and these are connected in series. If an alternating voltage source is applied across it then the resultant voltage of the L-C circuit

$V=V_{L} - V_{C} \qquad(1)$

We know that:

$V_{L} = iX_{L}$
$V_{C} = iX_{C}$

So from equation $(1)$

$V= iX_{L} - iX_{C} $

$V=i \left(X_{L} - X_{C} \right) $

$\frac{V}{i}=\left(X_{L} - X_{C} \right) $

$Z=\left(X_{L} - X_{C} \right) \qquad(2)$

Where
$Z \rightarrow$ Impedance of L-C circuit.
$X_{L} \rightarrow$ Inductive Reactance which has value $\omega L$
$X_{C} \rightarrow$ Capacitive Reactance which has value $\frac{1}{\omega C}$

So from equation $(2)$, we get

$Z=\left( \omega L - \frac{1}{\omega C} \right) \qquad(3)$

The phase of resultant voltage:

The phase of resultant voltage from current is $90^{\circ}$ as shown in the figure above.

The Impedance and Phase at Resonance Condition:($X_{L} = X_{C}$):

At resonance $X_{L} = X_{C} \qquad(5)$

$\omega L = \frac{1}{\omega C}$

$\omega^{2} = \frac{1}{L C}$

$\omega = \sqrt{\frac{1}{L C}}$

$2 \pi f = \sqrt{\frac{1}{L C}}$

$ f = \frac{1}{2 \pi}\sqrt{\frac{1}{L C}}$

Where $f \rightarrow$ Natural frequency of the circuit.

1.) The Impedance of the circuit at resonance condition:

Substitute the resonance condition i.e. $X_{L} = X_{C}$ in equation $(2)$ then the impedance of the L-C Circuit

$Z=0$

The impedance of the L-C circuit at resonance condition is zero.

2.) The Phase of resultant voltage at resonance condition:

There is not any change in the phase of resultant voltage at resonance condition i.e. that will be the same $90^{\circ}$.

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