Calculate the critical angle of an optical fibre without cladding if the glass has a refractive index of 1.56.

40^o

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39.9^o

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30^o

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90^o

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Solution

The correct option is B
39.9^o

$sin$ c = $\frac{1}{n_{g l a s s}}$ = $\frac{1}{1.56}$ = 0.641

So, c = ${39.9}^{\circ}$

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Calculate the critical angle of an optical fibre without cladding if the glass has a refractive index of 1.56.

The correct option is B 39.9o sin c = 1nglass = 11.56 = 0.641 So, c = 39.9∘​

The correct option is B
39.9^o

$sin$ c = $\frac{1}{n_{g l a s s}}$ = $\frac{1}{1.56}$ = 0.641

So, c = ${39.9}^{\circ}$