# Problem: For the scenarios in which total internal reflection is possible, rank the scenarios on the basis of the critical angle, the angle above which total internal reflection occurs. At this angle, the refracted ray is at 90 degrees from the normal. Rank from largest to smallest. To rank items as equivalent, overlap them.ITEM A: n1benzene =1.50   n2water =1.33ITEM B: n1diamond =2.42 n2water =1.33ITEM C: n1diamond =2.42 n2air =1.00ITEM D: n1water =1.33 n2air =1.00

###### FREE Expert Solution

Critical angle:

$\overline{){{\mathbf{\theta }}}_{{\mathbf{c}}}{\mathbf{=}}{\mathbf{s}}{\mathbf{i}}{{\mathbf{n}}}^{\mathbf{-}\mathbf{1}}{\mathbf{\left(}}{{\mathbf{\eta }}}_{{\mathbf{2}}}{\mathbf{/}}{{\mathbf{\eta }}}_{{\mathbf{1}}}{\mathbf{\right)}}}$

θc,A = sin-1 (1.33/1.50) = 62.45°

###### Problem Details

For the scenarios in which total internal reflection is possible, rank the scenarios on the basis of the critical angle, the angle above which total internal reflection occurs. At this angle, the refracted ray is at 90 degrees from the normal. Rank from largest to smallest. To rank items as equivalent, overlap them.

ITEM A: n1benzene =1.50   n2water =1.33

ITEM B: n1diamond =2.42 n2water =1.33

ITEM C: n1diamond =2.42 n2air =1.00

ITEM D: n1water =1.33 n2air =1.00

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Our tutors have indicated that to solve this problem you will need to apply the Total Internal Reflection concept. You can view video lessons to learn Total Internal Reflection. Or if you need more Total Internal Reflection practice, you can also practice Total Internal Reflection practice problems.