For an electromagnetic wave incident from one medium to a second medium, total reflection takes place when (a) The angle of incidence is equal to the Brewster angle with E field perpendicular to the plane of incidence (b) The angle of incidence is equal to the Brewster angle with E field parallel

Question

For an electromagnetic wave incident from one medium to a second medium, total reflection takes place when

(a) The angle of incidence is equal to the Brewster angle with E field perpendicular to the plane of incidence

(b) The angle of incidence is equal to the Brewster angle with E field parallel to the plane of incidence

(c) The angle of incidence is equal to the critical angle with the wave moving from the denser medium to a rarer medium

(d) The angle of incidence is equal to the critical angle with the wave moving from a rarer medium to a denser medium

Accepted Answer

Total internal inflection takes place when incident angle is greater than critical angel [latex] heta c[/latex].

[latex] heta_{c}=\sin ^{-1}\left(\frac{n_{2}}{n_{1}} ight)=\sin ^{-1} \sqrt{\frac{\in_{r_{2}}}{\in_{n_{1}}}}[/latex]

For total internal reflection

[latex] heta_{i} \geq heta_{c}[/latex]

[latex] heta_{i} \geq \sin ^{-1}\left(\sqrt{\frac{\xi_{r_{2}}}{\xi_{r_{1}}}} ight)[/latex]

For total internal reflection to take place, the wave should move from a denser medium to a rarer medium and the angle of incidence should be greater than or equal to the critical angle.
Hence, the correct option is (c)

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