Question

Consider scenarios A to F in which a ray of light traveling in material 1 is incident onto the interface with material 2. (Figure 1)

Part B

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.

n1water=1.33, n2air=1.00 n1diamond=2.42, n2air=1.00 n1diamond=2.42, n2water=1.33 n1benzene=1.50, n2water=1.33

Answer 1

$$ \begin{aligned} &n_{1} \sin \theta_{i}=n_{2} \sin \theta_{t} \\ &\sin \theta_{i}=\frac{n_{2}}{n_{1}} \sin \theta_{t} \end{aligned} $$

for critical angle, the value for \(\theta_{i}\) when \(\theta_{t}=90 \) and thus \(\sin \theta_{t}=1\).

\(\theta_{c}=\theta_{i}=\sin ^{-1}\left(\frac{n_{2}}{n_{1}}\right)\) n1water=1.33, n2air=1.00 \(\theta_{c}=\sin ^{-1}\left(\frac{1}{1.33}\right)=48.75^{\circ}\) n1diamond \(=2.42, n 2\) air \(=1.00\) \(\theta_{c}=\sin ^{-1}\left(\frac{1}{2.42}\right)=24.41^{\circ}\) n1diamond \(=2.42, n 2 w a t e r=1.33\) \(\theta_{c}=\sin ^{-1}\left(\frac{1.33}{2.42}\right)=33.34^{\circ}\) n1benzene \(=1.50, n 2 w a t e r=1.33\) \(\theta_{c}=\sin ^{-1}\left(\frac{1.33}{1.50}\right)=62.46^{\circ}\) Ranking 1)n1benzene \(=1.50, n 2 w a t e r=1.33\) 2)n1water \(=1.33, n 2\) air \(=1.00\) 3)n1diamond \(=2.42, n 2 w a t e r=1.33\) 4)n1diamond \(=2.42, n 2 a i r=100\)

1) 1 1 benzene \(=1.50, \mathrm{n} 2\) water \(=1.33\)

2) \(\mathrm{n} 1\) water \(=1.33, \mathrm{n} 2 \mathrm{air}=1.00\)

\(3) \mathrm{n} 1\) diamond \(=2.42, \mathrm{n} 2\) water \(=1.33\)

4) \(n\) 1diamond \(=2.42\), \(n 2 \mathrm{air}=100\)