Trace the rays of light showing the formation of an image due to a point object placed on the axis of a spherical surface separating the two media of refractive indices $n_1$ and $n_2$. Establish the relation between the distances of the object, the image and the radius of curvature from the central point of the spherical surface. Hence derive the expression of the lens maker’s formula.
CBSE DELHI - SET 1 2009
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$\tan<\text{NOM} = \frac{\text{MN}}{\text{OM}}$
$<\text{NCM} \cong \tan<\text{NCM} =\frac{\text{MN}}{\text{MC}}$
$<\text{NCM}\cong \tan<\text{NCM} =\frac{\text{MN}}{\text{MI}}$
Now, for $\bigtriangleup\text{NOC}$, i is the exterior angle, Therefore, $\text{I} = <\text{NOM} + <\text{NCM}$
$\ell = \frac{\text{MN}}{\text{OM}} + \frac{\text{MN}}{\text{MC}}$
Similary
$\text{r} = < \text{NCM} -<\text{NIM}'$$\text{i}.\text{e}..\text{r} =\frac{\text{MN}}{\text{MC}} - \frac{\text{MN}}{\text{MI}}$
Now , by Snell's law
$\text{n}_{1}\sin\text{i} = \text{n}_{2}\sin \text{r} , $
or for small angles
$\text{n}_{1}\text{i} = \text{n}_{2}\text{r}$
Substituting for i and r , we then get
$\frac{\text{n}_{1}}{\text{OM}} + \frac{\text{n}_{2}}{\text{MI}} = \frac{\text{n}_{2} - \text{n}_{1}}{\text{MC}}$
Applying the Cartesian sign convention,
$\text{MO} = - u, \text{MI} = + \text{v} , \text{MC} = + \text{R} , $ we get
$\frac{\text{n}_{2}}{v} - \frac{\text{n}_{1}}{u} = \frac{\text{n}_{2} - \text{n}_{1}}{\text{R}}$
Lens Maker’s Formula:
Applying the above result to the first interface ABC, we get,
$\frac{\text{n}_{1}}{\text{OB}} + \frac{\text{n}_{2}}{\text{BI}_{1}} = \frac{\text{n}_{2} - \text{n}_{2}}{\text{BC}_{1}}$
A similar procedure applied to the second interfaceADC gives,
$ - \frac{\text{n}_{2}}{\text{DI}_{1}} + \frac{\text{n}_{1}}{\text{DI}} = \frac{\text{n}_{2} - \text{n}_{2}}{\text{DC}_{2}}$
For a thin lens, $BI_1 = DI_1$. Adding the above two equations,we then get,
$\frac{\text{n}{1}}{\text{OB}} + \frac{\text{n}_{1}}{\text{DI}} = \big(\text{n}_{2} - \text{n}_{1}\big)\bigg(\frac{1}{\text{BC}_{1}} + \frac{1}{\text{DC}_{2}}\bigg)$
Suppose the object is at infinity. Then
$\text{OB} = \infty$ and $\text{DI} = f$ and the above equation becomes
$\frac{\text{n}_{1}}{f} = \big(\text{n}_{2} - \text{n}_{1}\big)\bigg(\frac{1}{\text{BC}_{1}} + \frac{1}{\text{DC}_{2}}\bigg)$
By the sign convention,
$\text{BC}_{1} = + \text{R}_{1}.$
$\text{DC}_{1} = - \text{R}_{2}$ $\frac{1}{f} = \big(\text{n}_{21} - 1\big)\bigg(\frac{1}{\text{R}_{1}} - \frac{1}{\text{R}_{2}}\bigg)$ $\big(\because\text{n}_{21} = \frac{\text{n}_{2}}{\text{n}_{1}}\big)$
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