Light ray incident along a vector $\overrightarrow{AO}$ ($\overrightarrow{AO} = 2\hat{i} - 3\hat{j}$) emerges out along vector $\overrightarrow{OB}$ ($\overrightarrow{OB} = C\hat{i} - 4\hat{j}$) as shown in the figure below. The value of $C$ is ______.
Answer: (A) $1.6$
Concept. The boundary is horizontal, so the normal is along $\hat j$ and the sine of each angle with the normal is $\dfrac{|x\text{-component}|}{\text{length}}$. Then apply Snell's law.
Step 1. Sines. $\sin\alpha=\dfrac{2}{\sqrt{13}}$; $\sin\beta=\dfrac{C}{\sqrt{C^2+16}}$.
Step 2. Snell's law. $1\times\dfrac2{\sqrt{13}}=1.5\dfrac{C}{\sqrt{C^2+16}}$. Squaring: $\dfrac4{13}(C^2+16)=2.25C^2$, so $4C^2+64=29.25C^2$.
Step 3. Solve. $C^2=\dfrac{64}{25.25}=2.53$, $C=1.59\approx1.6$.
Answer. $1.6$ (Option A).
Solution by Sreeraj P, M.Sc Physics
