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Explain internal reflection and total internal reflection. |
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Answer» Solution : When light travels from an optically denser medium to a rarer medium at the interface, it is partly REFLECTED back into the same medium and partly refracted to the second medium. This reflection is called the internal reflection. When a ray of light enters from a denser medium to a rarer medium it bends away from the normal.  The ray `AO_1B` in FIGURE the incident ray `AO_1`, is partially reflected `(O_1C)` and partially transmitted `(O_1B)` or refracted. The angle of refraction (r) being larger than the angle of incidence (i). As the angle of incidence increases, so does the angle of refraction, till for the ray `AO_3`, the angle of refraction is `pi/2` The refracted ray is bent so much away from the normal that it grazes the surface at the interface between the two media. This is shown by the ray `AO_3D`. The angle of incidence for which the refraction angle becomes `pi/2`, that incidence angle is called critical angle `.i_C.`. The deviated ray obtained after incidence at critical angle is called critical ray. If the angle of incidence is increased still further e.g. the ray `AO_4`, refraction is not possible and the incident ray is TOTALLY reflected. This is called total internal reflection, When light GETS reflected by a surface, normal some fraction of it gets transmitted. The reflected ray, therefore, is always less inten: than the incident ray, howsoever smooth th reflecting surface may be. In total intern reflection on the other hand, no transmission light takes place. According to Snell.s law at point `O_3`, `nisin i_c = n_2 sin r` `therefore (n_1)/(n_2)sin i_c = sin" "r` `therefore n_(12) sin i_c = sin 90^@" " [because r = 90^@]` `therefore n_(12)sin i_c = 1` `therefore _(12) = (1)/(sin i_c)`(where `n_(12)` = refractive inde of denser medium-1 w.r.t. rarer medium-2) If `n_1 = n` and `n_2` = 1 (air) then, sin `i_c` = 1 `therefore n=(1)/(sin i_c)` |
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