A point object 'O' is kept in a medium of refractive index $n_1$ in front of a convex spherical surface of radius of curvature R which separates the second medium of refractive index $n_2$ from the first one, as shown in the figure. Draw the ray diagram showing the image formation and deduce the relationship between the object distance and the image distance in terms of $n_1, n_2$ and R.
- hen the image formed above acts as a virtual object for concave spherical, surface separating the medium $n_2$ from $n_1 (n_2> n_1),$ draw this ray diagram and write the similar (similar to (a) relation. Hence obtain the expression for the lens maker's formula.
CBSE DELHI - SET 1 2015
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For small angles
$\angle\text{N0M}\cong\tan\angle \text{N0M} = \frac{\text{MN}}{\text{OM}}$
$\angle\text{NCM}\cong\tan\angle \text{NCM} = \frac{\text{MN}}{\text{MC}}$
$\angle\text{NIM}\cong\tan\angle \text{NIM} = \frac{\text{MN}}{\text{MI}}$
In $\Delta\text{NOC} , $ $\angle \text{i} = \angle\text{i}\angle\text{NOM} + \angle\text{NCM}$...........................(i)
$\therefore\angle\text{i} = \frac{\text{MN}}{\text{OM}} + \frac{\text{MN}}{\text{MC}}$
$\angle\text{r} = \angle \text{NCM} -\angle \text{NIM}$
$ =\frac{\text{MN}}{\text{MC}} -\frac{\text{MN}}{\text{MI}}$
Using Snell's Law
$n_1 \sin\ i. = n_2 \sin\ r$
For small angles $n_1 i = n_2 r$
Substituting for i and r, we get
$\frac{\text{n}_{1}}{\text{OM}} +\frac{\text{n}_{2}}{\text{MI}} = \frac{\text{n}_{2} -\text{n}_{1}}{\text{MC}}$
Here, $\text{OM} =\text{-u}, \text{MI} = + \text{v},\text{MC} = + \text{R}$
Substituting these, we get
$\Rightarrow\frac{\text{n}_{2}}{\text{v}} - \frac{\text{n}_{1}}{\text{u}} = \frac{\text{n}_{2} -\text{n}_{1}}{\text{R}}$
(Alternatively accept this Ray diagram)
Similarly relation for the surface ADC.
$\frac{-\text{n}}{\text{DI}{1}} +\frac{\text{n}_{1}}{\text{DI}} = \frac{\text{n}_{2} - \text{n}_{1}}{\text{DC}_{2}}$................................(i)
Refraction at the first surface ABC of the lens.
$\frac{\text{n}_{1}}{\text{OB}} +\frac{\text{n}_{2}}{\text{BI}_{1}} = \frac{\text{n}_{2} -\text{n}_{2}}{\text{BC}_{1}}$..............................(ii)
Adding (i)and (ii), and taking $ BI_1\cong DI_1,$ we get
$\frac{\text{n}_{1}}{\text{OB}} +\frac{\text{n}_{1}}{\text{DI}} = (\text{n}_{2} -\text{n}_{1})\big(\frac{1}{\text{BC}_{1}} + \frac{1}{\text{DC}_{2}}\big)$
Here, $OB = – u$
$DI = + v$
$BC_1 = + R_1$
$DC_2 = – R_2$
$\Rightarrow = \frac{\text{n}_{1}}{\text{-u}} + \frac{\text{n}_{1}}{\text{v}} = (\text{n}_{2} -\text{n}_{1})\bigg(\frac{1}{\text{R}_{1}} +\frac{1}{\text{R}_{2}}\bigg)$
$\Rightarrow = \text{n}_{1}\bigg(\frac{1}{\text{v}} +\frac{1}{u}\bigg) (\text{n}_{2} -\text{n}_{1})\bigg(\frac{1}{\text{R}_{1}} +\frac{1}{\text{R}_{2}}\bigg)$
$\Rightarrow\frac{1}{\text{f}} =\bigg(\frac{\text{n}_{2}}{\text{n}_{1}} -1 \bigg)\bigg(\frac{1}{\text{R}_{1}} -\frac{1}{\text{R}_{2}}\bigg)$
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