| In the figure [a] the light is incident at an angle \[{{\mu }_{k}}=1-\frac{1}{{{n}^{2}}}\] (slightly greater than the critical angle). Now keeping the incident ray fixed a parallel slab of refractive index \[{{n}_{3}}\] is placed on surface AB. |
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A) total internal reflection occurs at AB for \[{{n}_{3}}={{n}_{2}}\]
B) total internal reflection occurs at AB for \[{{n}_{3}}>{{n}_{1}}\]
C) the ray will return back to the same medium for all values of \[{{n}_{3}}\]
D) total reflection occurs at CD for \[{{n}_{3}}<{{n}_{1}}\].
Correct Answer: C
Solution :
| [c] For combination of \[{{n}_{1}}\] and \[{{n}_{2}}\] critical angle is \[\theta \] |
| When be places parallel slab of refractive index \[{{n}_{3}}({{n}_{3}}<{{n}_{1}})\] |
| Then critical angle decreases and T.I.R. takes place. |
| For any value of \[{{n}_{3}}\] the ray will be return back in the medium \[{{n}_{2}}\] and reflection takes place at the surface CD. |
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