Three sheets of plastic have unknown indices of refraction. Sheet 1 is placed on top of sheet $2,$ , and a laser beam is directed onto the sheets from above. The laser beam enters sheet 1 and then strikes the interface between sheet 1 and sheet 2 at an angle of ${26.5}^{\circ}$ with the normal. The refracted beam in sheet 2 makes an angle of ${31.7}^{\circ}$ with the normal. The experiment is repeated with sheet 3 on top of sheet $2,$ and, with the same angle of incidence on the sheet $3 -$ sheet 2 interface, the refracted beam makes an angle of ${36.7}^{\circ}$ with the normal. If the experiment is repeated again with sheet 1 on top of sheet $3,$ with that same angle of incidence on the sheet $1 -$ sheet 3 interface, what is the expected angle of refraction in sheet $3 ?$

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Solution

For sheets 1 and 2 as described,
$\begin{matrix} n_{1} sin {26.5}^{\circ} = n_{2} sin {31.7}^{\circ} 0.849 n_{1} = n_{2} \end{matrix}$
For the trial with sheets 3 and 2 ,
$\begin{matrix} n_{3} sin {26.5}^{\circ} = n_{2} sin {36.7}^{\circ} 0.747 n_{3} = n_{2} \end{matrix}$
Equate the two expressions for $n_{2}$
$0.747 n_{3} = 0.849 n_{1}$
$n_{3} = 1.14 n_{1}$
For the third trial,
$\begin{matrix} n_{1} sin {26.5}^{\circ} = n_{3} sin θ_{3} = 1.14 n_{1} sin θ_{3} θ_{3} = {23.1}^{\circ} \end{matrix}$

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For sheets 1 and 2 as described,n1sin26.5∘=n2sin31.7∘0.849n1=n2For the trial with sheets 3 and 2 ,n3sin26.5∘=n2sin36.7∘0.747n3=n2Equate the two expressions for n20.747n3=0.849n1n3=1.14n1For the third trial,n1sin26.5∘=n3sinθ3=1.14n1sinθ3θ3=23.1∘