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Physics-

General

Easy

Question

Three right angled prisms of refractive indices $n subscript 1 end subscript comma n subscript 2 end subscript$ and $n subscript 3 end subscript$ are fixed together using an optical glue as shown in figure. If a ray passes through the prisms without suffering any deviation, then

$n_{1} = n_{2} = n_{3}$
$n_{1} = n_{2} \neq n_{3}$
$1 + n_{1} = n_{2} + n_{3}$
$1 + n_{2}^{2} = n_{1}^{2} + n_{3}^{2}$

The correct answer is: $1 plus n subscript 2 end subscript superscript 2 end superscript equals n subscript 1 end subscript superscript 2 end superscript plus n subscript 3 end subscript superscript 2 end superscript$

At B
$sin invisible function application i equals n subscript 1 end subscript sin invisible function application r subscript 1 end subscript$ Þ $sin to the power of 2 end exponent invisible function application i equals n subscript 1 end subscript superscript 2 end superscript sin to the power of 2 end exponent invisible function application r subscript 1 end subscript$ (i)
At C
$n subscript 1 end subscript sin invisible function application left parenthesis 90 minus r subscript 1 end subscript right parenthesis equals n subscript 2 end subscript sin invisible function application r subscript 2 end subscript$ Þ $n subscript 2 end subscript superscript 2 end superscript sin to the power of 2 end exponent invisible function application r subscript 2 end subscript equals n subscript 1 end subscript superscript 2 end superscript cos to the power of 2 end exponent invisible function application r subscript 1 end subscript$ (ii)
At D
$n subscript 2 end subscript sin invisible function application left parenthesis 90 minus r subscript 2 end subscript right parenthesis equals n subscript 3 end subscript sin invisible function application r subscript 3 end subscript$ Þ $n subscript 2 end subscript superscript 2 end superscript cos to the power of 2 end exponent invisible function application r subscript 2 end subscript equals n subscript 3 end subscript superscript 2 end superscript sin to the power of 2 end exponent invisible function application r subscript 3 end subscript$ (iii)
At E
$n subscript 3 end subscript sin invisible function application left parenthesis 90 minus r subscript 3 end subscript right parenthesis equals left parenthesis 1 right parenthesis sin invisible function application left parenthesis 90 minus 1 right parenthesis$ Þ $cos to the power of 2 end exponent invisible function application i equals n subscript 3 end subscript superscript 2 end superscript cos to the power of 2 end exponent invisible function application r subscript 3 end subscript$ (iv)
Adding (i), (ii), (iii) and (iv) we get $1 plus n subscript 2 end subscript superscript 2 end superscript equals n subscript 1 end subscript superscript 2 end superscript plus n subscript 3 end subscript superscript 2 end superscript$

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