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Physics-

General

Easy

Question

An optical fibre consists of core of $mu subscript 1 end subscript$ surrounded by a cladding of $mu subscript 2 end subscript less than mu subscript 1 end subscript$ . A beam of light enters from air at an angle $alpha$ with axis of fibre. The highest $alpha$ for which ray can be travelled through fibre is

$\cos^{- 1} \sqrt{μ_{2}^{2} - μ_{1}^{2}}$
$\sin^{- 1} \sqrt{μ_{1}^{2} - μ_{2}^{2}}$
$\tan^{- 1} \sqrt{μ_{1}^{2} - μ_{2}^{2}}$
$\sec^{- 1} \sqrt{μ_{1}^{2} - μ_{2}^{2}}$

The correct answer is: $sin to the power of negative 1 end exponent invisible function application square root of mu subscript 1 end subscript superscript 2 end superscript minus mu subscript 2 end subscript superscript 2 end superscript end root$

Here the requirement is that $i greater than c$
$rightwards double arrow sin invisible function application i greater than sin invisible function application c rightwards double arrow sin invisible function application i greater than fraction numerator mu subscript 2 end subscript over denominator mu subscript 1 end subscript end fraction$ (i)
From Snell’s law $mu subscript 1 end subscript equals fraction numerator sin invisible function application alpha over denominator sin invisible function application r end fraction$ (ii)

Also in $capital delta O B A$
$r plus i equals 9 0 to the power of o end exponent$ $rightwards double arrow r equals left parenthesis 90 minus i right parenthesis$
Hence from equation (ii)
$sin invisible function application alpha equals mu subscript 1 end subscript sin invisible function application left parenthesis 90 minus i right parenthesis$
$rightwards double arrow cos invisible function application i equals fraction numerator sin invisible function application alpha over denominator mu subscript 1 end subscript end fraction$
$sin invisible function application i equals square root of 1 minus cos to the power of 2 end exponent invisible function application i end root equals square root of 1 minus open parentheses fraction numerator sin invisible function application alpha over denominator mu subscript 1 end subscript end fraction close parentheses to the power of 2 end exponent end root$ ….(iii)
From equation (i) and (iii) $square root of 1 minus open parentheses fraction numerator sin invisible function application alpha over denominator mu subscript 1 end subscript end fraction close parentheses to the power of 2 end exponent end root greater than fraction numerator mu subscript 2 end subscript over denominator mu subscript 1 end subscript end fraction$
Þ $sin to the power of 2 end exponent invisible function application alpha less than left parenthesis mu subscript 1 end subscript superscript 2 end superscript minus mu subscript 2 end subscript superscript 2 end superscript right parenthesis$ Þ $sin invisible function application alpha less than square root of mu subscript 1 end subscript superscript 2 end superscript minus mu subscript 2 end subscript superscript 2 end superscript end root$
$alpha sin to the power of negative 1 end exponent invisible function application square root of mu subscript 1 end subscript superscript 2 end superscript minus mu subscript 2 end subscript superscript 2 end superscript end root subscript m a x end subscript$

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