Physics-
General
Easy

Question

A rectangular glass slab ABCD of refractive index n1 is immersed in water of refractive index n subscript 2 end subscript open parentheses n subscript 1 end subscript greater than n subscript 2 end subscript close parentheses A ray of light is incident a the surface AB of the slab as shown The maximum value of the angle of incidence amax, such that the ray comes out only from the other surface CD, is given by

  1. s i n to the power of negative 1 end exponent invisible function application open square brackets fraction numerator n subscript 1 end subscript over denominator n subscript 2 end subscript end fraction c o s invisible function application open parentheses sin to the power of negative 1 end exponent invisible function application fraction numerator n subscript 2 end subscript over denominator n subscript 1 end subscript end fraction close parentheses close square brackets    
  2. s i n to the power of negative 1 end exponent invisible function application open square brackets n subscript 1 end subscript c o s invisible function application open parentheses sin to the power of negative 1 end exponent invisible function application fraction numerator 1 over denominator n subscript 2 end subscript end fraction close parentheses close square brackets    
  3. s i n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator n subscript 1 end subscript over denominator n subscript 2 end subscript end fraction close parentheses    
  4. s i n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator n subscript 2 end subscript over denominator n subscript 1 end subscript end fraction close parentheses    

The correct answer is: s i n to the power of negative 1 end exponent invisible function application open square brackets fraction numerator n subscript 1 end subscript over denominator n subscript 2 end subscript end fraction c o s invisible function application open parentheses sin to the power of negative 1 end exponent invisible function application fraction numerator n subscript 2 end subscript over denominator n subscript 1 end subscript end fraction close parentheses close square brackets

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A point source S is placed at the bottom of different layers as shown in the figure The refractive index of bottom most layer is m0 The refractive index of any other upper layer is mu subscript left parenthesis n right parenthesis end subscript equals mu subscript 0 end subscript minus fraction numerator mu subscript 0 end subscript over denominator 4 n minus 18 end fraction text  where  end text n equals 1 comma 2 comma horizontal ellipsis horizontal ellipsis A ray of light with angle i slightly more than 30° starts from the source S Total internal reflection takes place at the upper surface of a layer having n equal to

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If A equals open square brackets table row 1 cell negative 2 end cell row 5 3 end table close square brackets, then A plus A to the power of T end exponentequals

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The inverse of the matrix open square brackets table row 3 cell negative 2 end cell row 1 4 end table close square bracketsis

Let A equals open square brackets table row 3 cell negative 2 end cell row 1 4 end table close square brackets rightwards double arrow vertical line A vertical line equals 14
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Let A equals open square brackets table row 3 cell negative 2 end cell row 1 4 end table close square brackets rightwards double arrow vertical line A vertical line equals 14
therefore a d j A equals open square brackets table row 4 2 row cell negative 1 end cell 3 end table close square brackets rightwards double arrow A to the power of negative 1 end exponent equals open square brackets table row cell fraction numerator 4 over denominator 14 end fraction end cell cell fraction numerator 2 over denominator 14 end fraction end cell row cell fraction numerator negative 1 over denominator 14 end fraction end cell cell fraction numerator 3 over denominator 14 end fraction end cell end table close square brackets.
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The ad joint of open square brackets table row 1 1 1 row 1 2 cell negative 3 end cell row 2 cell negative 1 end cell 3 end table close square bracketsis

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rightwards double arrow A subscript 11 end subscript equals 3 comma A subscript 12 end subscript equals negative 9 comma A subscript 13 end subscript equals negative 5
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rightwards double arrow A subscript 11 end subscript equals 3 comma A subscript 12 end subscript equals negative 9 comma A subscript 13 end subscript equals negative 5
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