For total internal reflection to take place- the angle of incidence i and the refractive index - of the medium must satisfy the inequalityA- 1-sin i-C- sin i -D- sin i -

The correct option is A 1/sin i lt; μ For total internal refection to take place i gt;θ_C or, sin i gt; sin θ_C We know that, sin θ_C = 1/Refractive in ...

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Standard X

Physics

Optical Fiber

For total int...

Question

For total internal reflection to take place, the angle of incidence $i$ and the refractive index ( $μ$ ) of the medium must satisfy the inequality

$\frac{1}{s i n i} < μ$

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$\frac{1}{s i n i} > μ$

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$s i n i < μ$

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$s i n i > μ$

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Solution

The correct option is A
$\frac{1}{s i n i} < μ$

For total internal refection to take place
$i > θ_{C}$
or, $s i n i > s i n θ_{C}$
We know that, $sin θC=1Refractive index of the medium=1μ$
So, $s i n i > \frac{1}{μ}$
or, $\frac{1}{s i n i} < μ$

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For total internal reflection to take place, the angle of incidence i and the refractive index (μ) of the medium must satisfy the inequality

The correct option is A 1sin i<μ For total internal refection to take place i>θC or, sin i>sin θC We know that, sin θC=1Refractive index of the medium=1μ So, sin i>1μ or, 1sin i<μ

For total internal reflection to take place, the angle of incidence $i$ and the refractive index ( $μ$ ) of the medium must satisfy the inequality

The correct option is A
$\frac{1}{s i n i} < μ$

For total internal refection to take place
$i > θ_{C}$
or, $s i n i > s i n θ_{C}$
We know that, $sin θC=1Refractive index of the medium=1μ$
So, $s i n i > \frac{1}{μ}$
or, $\frac{1}{s i n i} < μ$