Question

(a) Derive the mathematical relation between refractive indices $n_1$ and $n_2$ of two media and radius of curvature R for refraction at a convex spherical surface. Consider the object to be a point since lying on the principle axis in rarer medium of refractive index $n_1$ and a real image formed in the denser medium of refractive index $n_2$. Hence, derive Lens Maker's formula.

(b) Light from a point source in air falls on a convex spherical glass surface of refractive index 1.5 and radius of curvature 20 cm. The distance of light source from the glass surface is 100 cm. At what position is the image formed ?

Answer 1

(a) $tan\angle NOM=\frac{MN}{OM};$

$tan\angle NCM=\frac{MN}{MC};$

$tan\angle NIM=\frac{MN}{MI};$

For $△NOC$, i is the exterior angle.

Assuming the incident ray is very close to the principal axis, all the angles are very small. Hence, for very small angles,

$tanx=x=sinx$

$\therefore i=\angle NOM+\angle NCM$

$i=\frac{MN}{OM}+\frac{MN}{MC}...\left(i\right)$

Similarly,$r=\angle NCM-\angle NIM$

i.e., $r=\frac{MN}{MC}-\frac{MN}{MI}...\left(ii\right)$

By Snell's law,
$n_{1}sini=n_{2}sinr$

For small angles,
$n_{1}i=n_{2}r$

On Substituting these values of i and r in equations, we get

$n_1(\frac{MN}{OM}+\frac{MN}{MC})=n_2(\frac{MN}{MC}-\frac{MN}{MI})$

$\frac{n_1}{OM}+\frac{n_2}{MI}=\frac{n_2-n_1}{MC}...\left(iii\right)$

On applying new Cartesian sign conventions, $OM=-u,MI=+v,MC=+R$

Substituting these values in equation (iii),we get

$\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}...\left(iv\right)$

In deriving Lens maker's formula, we adopt the coordinate geometry sign convention and make the assumptions :

(i) The lens is so thin, that the distances measured from the poles of its two surfaces can be taken as equal to the distances from its optical centre.

(ii) The aperture of the lens is small.

(iii)The object is a point-object placed on the principal axis of the lens.

(iv) The incident and the refracted rays make small angles with the principal axis.


Refraction at the first surface,
$\frac{n_2}{v^{\prime }}-\frac{n_1}{u}=\frac{n_2-n_1}{R_1}$

Refraction at the second surface,
$\frac{n_1}{v}-\frac{n_2}{v^{\prime }-t}=\frac{n_1-n_2}{R_2}$

The lens is 'thin', hence t < < v' and can be ignored. Then, we have

$\frac{n_1}{v}-\frac{n_2}{v^{\prime }}=\frac{n_1-n_2}{R_2}...\left(ii\right)$

Adding equation (i) and (ii), we get

$\frac{n_1}{v}-\frac{n_2}{u}=n_2-n_1(\frac{1}{R_1}-\frac{1}{R_2})$

Putting $\frac{n_2}{n_1}=n$,the refractive index of the material of the lens with respect to the surrounding medium, we have

$\frac{1}{v}-\frac{1}{u}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})...\left(iii\right)$

When the object is at infinity, the image will be formed at the principal focus of the lens, i.e., when $u=\infty$,

$\frac{1}{f}-\frac{1}{\infty }=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\frac{1}{f}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})$

This is the Lens Maker's formula.

(b) Given,refractive index,$n_2=1.5,n_1=1\left(air\right)$

Radius of curvature,R=20 cm

Object distance,u=-100 cm

To find,Image distance,v

$\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$

$\frac{1.5}{v}+\frac{1}{\left(100\right)}=\frac{1.5-1}{20}=\frac{1}{40}$

$\frac{1.5}{v}=\frac{1}{40}-\frac{1}{100}=\frac{5-2}{200}=\frac{3}{200}$

$v=\frac{200}{3}\times 1.5=100cm$

The image is formed at 100 cm in denser medium.