- Derive the mathematical relation between refractive indices $n_1$ and $n_2 $ of two radii 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.
- 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?
CBSE OUTSIDE DELHI - SET 3 CENTRAL 2016
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$\theta_{1} =\alpha + \beta$
$\theta_{2} = \beta - \gamma$
$\therefore \gamma = \beta - \theta $
For paraxial rays $\theta_{1}$ and $\theta_{2}$ are small Therefore,$ n_2 \sin \theta_{2} = n_1 \sin \theta_{2}$ (Snells law)
Reduces to,
At N $\frac{\sin\text{i}}{\sin\text{r}}\sim\frac{\text{i}}{\text{r}} = \frac{\text{n}_{2}}{\text{n}_{1}}$
$\therefore\text{n}_{1} =\text{rXn}_{2}$
$( \alpha +\beta ) \text{n}_{1} = (\beta - \theta ) \text{n}_{2}$
$\text{n}_{1}\big(\frac{\text{NM}}{\text{OM}} + \frac{\text{NM}}{\text{MC}}\big) = \big(\frac{\text{NM}}{\text{MC}} - \frac{\text{NM}}{\text{MI}}\big)\text{n}_{2}$
$\text{n}_{1}\big(\frac{1}{\text{ - u}} +\frac{1}{\text{ + R }}\big) = \big(\frac{1}{\text{ + R}} - \frac{1}{\text{u}}\big)\text{n}_{2}$
$\frac{\text{n}_{2}}{\text{v}'} - \frac{\text{n}_{1}}{\text{u}} = \frac{(\text{n}_{2} - \text{n}_{1})}{\text{R}_{1}}$
Appying above relations to refraction through a lens:
$\frac{\text{n}_{2} - \text{n}_{1}}{\text{R}_{1}} = \frac{\text{n}_{1}}{\text{v}'} - \frac{\text{n}_{1}}{\text{u}}$ . ... . ... .. (i)
For surface 2
$\frac{\text{n}_{1} - \text{n}_{2}}{\text{R}_{2}} = \frac{\text{n}_{1}}{\text{v}} - \frac{\text{n}_{2}}{\text{v}'}$ . . .. . . . . (ii)
Adding eqn. (i) and (ii)
$(\text{n}_{2} - \text{n}_{1}) \big(\frac{1}{\text{R}_{1}} -\frac{1}{\text{R}_{2}}\big) = \text{n}_{1}\big(\frac{1}{\text{v}} - \frac{1}{\text{u}}\big)$
For $\text{u}=\propto \text{v}=\text{f}$
$\therefore \frac{\text{n}_{1}}{\text{f}} = (\text{n}_{2} - \text{n}_{1})\big(\frac{1}{\text{R}_{1}} - \frac{1}{\text{R}_{2}}\big)$
$\Rightarrow\frac{1}{\text{f}} = \bigg(\frac{\text{n}_{2}}{\text{n}_{1}} - 1 \bigg)\bigg(\frac{1}{\text{R}_{1}} - \frac{1}{\text{R}_{2}}\bigg)$
- $R = 20$ cm $n_2 = 1.5$ $n_1 = 1$ $u = - 100$ cm
$ = \frac{0.5}{20\text{cm}} - \frac{1}{100\text{cm}}$
$ =\frac{1.5}{100}\text{cm}$
$\Rightarrow\text{V} = 100 \text{ cm}$ a real image on the other side, 100 cm away from the surface.
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