Question
Explain the refraction of light from a denser medium to a rarer medium, using Huygen's principle.
✓
Answer
→As shown in the fig., a wavefront enters from a denser to rarer medium.
→Speed of light in the denser and rarer medium are $v_1$ and $v_2$ and the refractive indices of the medium 1 and 2 are $n_1$ and $n_2$, respectively. $\left(n_1>n_2\right)$
→As shown in fig., a wavefront AB is incident at an angle $i$ on a surface separating medium 1 and 2 .
(i) $i→Figure for this situation is shown above. Where CE is the refracted wavefront. Snell's law $n_1 \sin i=n_2 \sin r$ is followed in this condition.
→Here, since $n_1>n_2$ the angle of incidence $(r>i)$ is less then the angle of refrection.
If $i=i_c$ (Angle of incidence is equal to Critical angle), $ r=90^{\circ} \text {. } $
→$\therefore$ From Snell's law, $ \sin i_c=\frac{n_2}{n_1} $
(ii) $i>i_c$ (Angle of incidence is greater than critical angle)
→If the angle of incidence is greater than the critical angle, we will not have any refracted wave and the wave will undergo what is known as total internal reflection.
→Speed of light in the denser and rarer medium are $v_1$ and $v_2$ and the refractive indices of the medium 1 and 2 are $n_1$ and $n_2$, respectively. $\left(n_1>n_2\right)$
→As shown in fig., a wavefront AB is incident at an angle $i$ on a surface separating medium 1 and 2 .
(i) $i
→Here, since $n_1>n_2$ the angle of incidence $(r>i)$ is less then the angle of refrection.
If $i=i_c$ (Angle of incidence is equal to Critical angle), $ r=90^{\circ} \text {. } $
→$\therefore$ From Snell's law, $ \sin i_c=\frac{n_2}{n_1} $
(ii) $i>i_c$ (Angle of incidence is greater than critical angle)
→If the angle of incidence is greater than the critical angle, we will not have any refracted wave and the wave will undergo what is known as total internal reflection.
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