Light ray incident along a vector $$\vec{AO}$$ ($$\vec{AO} = 2\hat{i} - 3\hat{j}$$) emerges out along vector $$\vec{OB}$$ ($$\vec{OB} = C\hat{i} - 4\hat{j}$$) as shown in the figure below. The value of C is __________.

Use Snell’s law:

n₁ sinθ₁ = n₂ sinθ₂
here n₁ = 1, n₂ = 1.5

Angles are measured w.r.t. the normal (vertical axis).

So use horizontal components of direction vectors.

Incident ray AO = (2, −3)

$$\sin\theta_1=\frac{|x|}{\sqrt{x^2+y^2}}=\frac{2}{\sqrt{13}}$$

Refracted ray OB = (C, −4)

$$\sin\theta_2=\frac{|C|}{\sqrt{C^2+16}}$$

Apply Snell’s law:

$$\frac{2}{\sqrt{13}}=1.5\cdot\frac{C}{\sqrt{C^2+16}}$$

Solve:

$$\frac{2}{\sqrt{13}}=\frac{3}{2}\cdot\frac{C}{\sqrt{C^2+16}}$$

$$\frac{4}{\sqrt{13}}=\frac{3C}{\sqrt{C^2+16}}$$

Square:

$$\frac{16}{13}=\frac{9C^2}{C^2+16}$$

$$16(C^2+16)=117C^2$$

$$16C^2+256=117C^2$$

$$101C^2=256$$

$$C^2=\frac{256}{101}$$

$$C=\frac{16}{\sqrt{101}}\approx1.6$$