Thin Lens and Mirrors

THIN LENS

Consider the thin lens shown here with the two refracting surfaces having radii of curvature equal to R₁ and R₂ respectively. The refractive indices of this surrounding medium and of the material of the lens are $\mu$₁ and $\mu$₂ respectively.

Now using the result that we obtained for refraction at single spherical surface we get,

For first surface, $\frac{{\mu }_2}{v_1}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R_1}$ ...(1)

For second surface, $\frac{{\mu }_1}{v}-\frac{{\mu }_1}{v_1}=\frac{{\mu }_1-{\mu }_2}{R_2}$ ...(2)

Adding (1) and (2), $\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\left({\mu }_1-{\mu }_2\right)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$

$\Rightarrow \frac{1}{v}-\frac{1}{u}=\left(\frac{{\mu }_2}{{\mu }_1}-1\right)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$

When u = $\infty$, v = f, $\frac{1}{f}=\left(\frac{{\mu }_2}{{\mu }_1}-1\right)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$ Lensmaker's Formula

Also, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ Thin Lens Formula

EQUIVALENT FOCAL LENGTH OF TWO OR MORE THIN LENSES IN CONTACT

$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}+...$

Special cases

(a) If the lens is immersed in a liquid whose refractive index is greater than refractive index of material of the lens then

$\frac{1}{f_{(liquid)}}={(}_I{\mu }_g-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$

f_liquid is the focal length of lens in the liquid _I$\mu$_g is the R.1. of glass w.r.t. liquid. The focal length of the lens in the liquid becomes negative.

(b) If the lens is immersed in a liquid whose R.1. is equal to the R.1. of the material of the lens then

${}_I{\mu }_g=\frac{{}_a{\mu }_b}{{}_a{\mu }_I}=1$

$\therefore$ $\frac{1}{f_{(liquid)}}=(1-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]=0$

The focal length becomes infinite

(c) If the lens be immersed in a liquid whose R.1. is less than the R.1. of the material of the lens

$\frac{1}{f_{(liquid)}}={(}_I{\mu }_g-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$

The focal length of the lens increases.

In general, if $\mu$ decreases, f increases

Note: A lens is converging if its focal length is positive and diverging if focal length is negative (in neo-Cartesian sign convention). From this we can conclude that a convex lens need not necessarily be a converging and a concave lens diverging.

Ray tracing for lens

Illustration 1: A glass rod has ends as shown in figure, The refractive index of glass is $\mu$. The object point O is at a distance 2R from the surface of larger radius of curvature. The distance between the apexes of the ends is 3R. Show that the image point of O is formed at a distance of $\frac{\left(9-4\mu \right)R}{\left(10\mu -9\right)\left(\mu -2\right)}$

Solution: From the right hand vertex.

For the first surface

$\frac{{\mu }_2}{v^{\prime }}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R_1}$

Here $\mu$₂ =$\mu$, $\mu$₁ =1, $\mu$ = –2R, R₁ = +R

$\frac{u}{v^{\prime }}+\frac{1}{2R}=\frac{\mu -1}{+R}$

$\frac{u}{v^{\prime }}=\frac{\mu -1}{R}-\frac{1}{2R}=\frac{2\mu -3}{2R}$

$\therefore v^{\prime }=\frac{2\mu R}{2\mu -3}$

For the second surface

$u=-\left[3R-\left(\frac{2\mu R}{2\mu -3}\right)\right]$

Here $\mu$₂ = 1, $\mu$₁ = $\mu$, $R_2=+\frac{R}{2}$

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R_2}$

$\frac{1}{v}+\frac{\mu }{3R-\left[\begin{array}{c}2\mu R\\ 2\mu -3\end{array}\right]}=\frac{1-\mu }{R/2}$

$\frac{1}{v}=\frac{2-2\mu }{R}-\frac{\mu \left(2\mu -3\right)}{\left(6\mu R-9R-2\mu R\right)}$

$\frac{1}{v}=\frac{2-2\mu }{R}-\frac{\mu \left(2\mu -3\right)}{R\left[4\mu -9\right]}=\frac{-10{\mu }^2+29\mu -18}{R\left[4\mu -9\right]}$

$v=\frac{\left(9-4\mu \right)R}{10{\mu }^2-29\mu +18}=\frac{\left(9-4\mu \right)R}{\left(10\mu -9\right)\left(\mu -2\right)}$

POWER OF A LENS

If f is in metres then the power P of the lens in dioptres is given by, $P=\frac{1}{f}$

If two lens in are separated by a distance d then power of combination of lens is

$P=P_1+P_2-\frac{d}{\mu }{P}_1{P}_2$

Where P₁ and P₂ are optical powers of the two lenses, and $\mu$ is the refractive index of the medium in between them.

Illustration 2: A lens has a power of +5 diopters in air. What will be its power if completely immersed in water?

${}_a{\mu }_w=4/3and{}_a{\mu }_g=3/2.$

Solution: Let f_a and f_w be the focal lengths of the lens in air and water respectively, then

$P_a=\frac{1}{f_a}or+5=\frac{1}{f_a}$

f_a = 0.2 m = 20 cm

Now $\frac{1}{f_a}=\left({}_a{\mu }_g-1\right)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$ ...(1)

and $\frac{1}{f_w}=\left({}_w{\mu }_g-1\right)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$ ...(2)

Dividing equation (1) by equation (2), we get,

$\frac{f_w}{f_a}=\left[\frac{{}_a{\mu }_g-1}{{}_w{\mu }_g-1}\right]$

Again, ${}_w{\mu }_g=\frac{{}_a{\mu }_g}{{}_w{\mu }_g}=\frac{3/2}{4/3}=\frac{9}{8}$

$\therefore \frac{f_w}{f_a}=\frac{\left(3/2\right)-1}{\left(9/8\right)-1}=\frac{\left(1/2\right)}{\left(1/8\right)}=4$

$f_w=f_a\times 4=20\times 4=80cm=0.8m$

$P_w=\frac{1}{f_w}=\frac{1}{0.8}=1.25diopter.$

SILVERING AT ONE SURFACE of lens

When one surface of a thin lens is silvered, then the focal length F of the effective lens-mirror combination is expressed as, $1/F=\sum 1/f_i$, where f_i is the focal length of the lens or mirror to be repeated as many times as the refraction or reflection respectively is repeated.

Some Cases:

(i) Focal length of plano–convex lens when silvered at its plane surface

When an object is placed in front of such a lens. The ray first of all are refracted from the convex surface, then reflected from the polished plane surface and again refracted out from the convex surface. If $f_{\ell }$ and f_m be the focal lengths of lens and mirror.

$\Rightarrow \frac{1}{F}=\frac{1}{f_{\ell }}+\frac{1}{f_m}+\frac{1}{f_{\ell }},Here,\frac{1}{f_{\ell }}=(\mu -1)\left[\frac{1}{R}\right]and\frac{1}{f_m}=\frac{2}{R_m}=\frac{2}{\infty }=0$

$\Rightarrow \frac{1}{F}=\frac{2}{f_{\ell }}\Rightarrow F=\frac{f_{\ell }}{2}=\frac{R}{2(\mu -1)}$

(ii) Focal length of plano –convex lens when silvered at convex surface

$\Rightarrow \frac{1}{F}=\frac{1}{f_{\ell }}+\frac{1}{f_m}+\frac{1}{f_{\ell }},Here,\frac{1}{f_{\ell }}=(\mu -1)\left[\frac{1}{R}\right]and\frac{1}{f_m}=\frac{2}{R}$

$\Rightarrow \frac{1}{F}=\frac{2}{f_{\ell }}+\frac{2}{R}\Rightarrow \frac{1}{F}=2\left[\frac{(\mu -1)}{R}\right]+\frac{2}{R}\Rightarrow F=\frac{R}{2\mu }$

(iii) Focal length of convex lens where convex surface of radius R₂ is silvered

$\Rightarrow \frac{1}{F}=\frac{1}{f_l}+\frac{1}{f_m}+\frac{1}{f_{\ell }},Here,\frac{1}{f_{\ell }}=(\mu -1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]and\frac{1}{f_m}=\frac{2}{R_2}$

$\Rightarrow \frac{1}{F}=\frac{2}{f_{\ell }}+\frac{2}{R_2}\Rightarrow \frac{1}{F}=2\left[(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\right]+\frac{2}{R_2}$

Illustration 3: An object is 1 metre in front of the curved surface of a plano–convex lens whose flat surface is silvered. A real image is formed 120 cm in front of the lens. What is the focal length of the lens?

Solution:

Here u = 100 cm

v = 120 cm

$\frac{1}{F}=\frac{1}{v}+\frac{1}{u}=\frac{1}{120}+\frac{1}{100}=\frac{11}{600}$

Now $\frac{1}{F}=\frac{1}{F_L}+\frac{1}{F_M}+\frac{1}{F_L}$

Where F_L – focal length of the lens.

F_M – focal length of the mirror.

$=\frac{2}{F_L}+\frac{1}{F_M}\left(\because F_M=\infty \right)$

$\frac{1}{F}=\frac{2}{F_L}$

$\therefore \frac{11}{600}=\frac{2}{F_L}$

$F_L=\frac{1200}{11}=109.1cm$

ANGLE OF DEVIATION OF A RAY WHEN IT PASSES A LENS

O is the object and I is the image $\delta$ is the angle of deviation.

$\delta =\alpha +\beta \approx \tan \alpha +\tan \beta$

$=\frac{h}{-u}+\frac{h}{v}=h\left[\frac{1}{v}-\frac{1}{u}\right]$

$\Rightarrow \delta =\frac{h}{f}$

Determination of focal length of a concave mirror by u–v method

The relation between object distance u and the image v from the pole of the mirror is given by,

$\begin{array}{c}\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\end{array}$

Where f is the focal length of the mirror. The focal length of the concave mirror is obtained either from $\frac{1}{v}$ versus $\frac{1}{u}$ graph or from u–v graph.

$\frac{1}{u}$ and $\frac{1}{v}$ graph

When the image is real (of course only upon then it can be obtained on screen), the object lies between focus (F) and infinity. In such a situation u, v and f all are negative. Hence the mirror formula

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

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

or again, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

or, $\frac{1}{v}=-\frac{1}{u}+\frac{1}{f}$

Comparing with y = mx + c, the desired graph will be a straight line with slope –1 and intercept equal to $\frac{1}{f}.$

The corresponding $\frac{1}{v}$ versus $\frac{1}{u}$ graph is as shown in figure. The intercepts on the horizontal and vertical axes are equal. It is equal to $\frac{1}{f}$. A straight line OC at an angle 45^o with the horizontal axis intersects line AB at C. The coordinates of point C are $\left(\frac{1}{2f},\frac{1}{2f}\right).$

The focal length of the mirror can be calculated by measuring the coordinates of either of the points A, B or C.

u–v graph

The u–v graph comes out to be a hyperbola as shown in figure.

A line drawn at angle 45^o from the origin intersects the hyperbola at point C. The coordinates of point C are (2f, 2f).

The focal length of mirror can be calculated by measuring the coordinates of point C.

Determination of focal length of convex lens using u–v method

The relation between u, v and f for a convex lens is,

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

$\frac{1}{u}$ and $\frac{1}{v}$ graph

Using the proper sign convention, u is negative, u negative, v and f are positive. So we have,

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

Comparing with y = mx + c, graph between $\frac{1}{v}$ and $\frac{1}{u}$ is negative with slope –1 and intercept $\frac{1}{f}.$ The corresponding graph is as shown in figure.

Proceeding in the similar manner as discussed in case of a concave mirror the focal length of the lens can be calculated by measuring the coordinates of either of the points A, B and C.

u–v Graph

The u versus v graph is a hyperbola as shown in figure. By measuring the coordinates of point C whose coordinates are (2f, 2f) we can calculate the focal length of the lens.