- For a ray of light travelling from a denser medium of refractive index $n_1$ to a rarer medium of refractive index $n_2$ prove that $\frac{\text{n}_{2}}{\text{n}_{1}} = \sin \text{i}_{c} ,$ where $i_c$ is the critical angle of incidence for the media.
- Explain with the help of a diagram, how the above principle is used for transmission of video signals using optical fibres.
CBSE DELHI - SET 1 2008
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- (i)
- $\frac{\text{n}_{2}}{\text{n}_{1}} = \frac{\sin\text{i}}{\sin\text{r}}$
$\angle\text{r} = 90^{o}$
$\therefore$ $\frac{\sin \text{i}_{c}}{\sin90^{\circ}} = \frac{\text{n}_{2}}{\text{n}_{1}}$
or $\sin\text{i}_{c} = \frac{\text{n}_{2}}{\text{n}_{1}}$
- (i)
(ii) When a video signal is directed at one end of the fiber at a suitable angle, it undergoes repeated total internal reflections along the length of the fiber and comes out of it.
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