Physics-
General
Easy

Question

An optical fibre consists of core of mu subscript 1 end subscriptsurrounded 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

  1. cos to the power of negative 1 end exponent invisible function application square root of mu subscript 2 end subscript superscript 2 end superscript minus mu subscript 1 end subscript superscript 2 end superscript end root    
  2. 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    
  3. tan 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    
  4. sec 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    

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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