Question

Incident ray is along \(3\hat i-2\hat j\) and refracted ray is along \(c\hat i-4\hat j\). Find \(c\).

• A. \(1.6\)
• B. \(0.6\)
• C. \(2.6\)
• D. \(4\)

Answer 1

The Correct Option is C

Concept: According to Snell's law, \[ \mu_1 \sin i = \mu_2 \sin r \] Angle with the normal can be obtained using the cross product of the ray direction with the unit normal vector. Normal to the surface is along the \(y\)-axis.
Step 1:Find unit vector along incident ray. \[ \hat e = \frac{3\hat i - 2\hat j}{\sqrt{3^2+(-2)^2}} \] \[ \hat e = \frac{3\hat i - 2\hat j}{\sqrt{13}} \]
Step 2:Find unit vector along refracted ray. \[ \hat r = \frac{c\hat i - 4\hat j}{\sqrt{c^2 + 16}} \]
Step 3:Use Snell's law in vector form. \[ \mu_1 |\hat e \times \hat n| = \mu_2 |\hat r \times \hat n| \] where \(\hat n\) is unit vector along normal. Given \[ \mu_1 = 1, \qquad \mu_2 = \frac{3}{2} \]
Step 4:Substitute values. \[ \frac{3}{\sqrt{13}} = \frac{3}{2}\frac{c}{\sqrt{c^2+16}} \]
Step 5:Solve the equation. \[ \frac{3}{\sqrt{13}+2^2} = \frac{3c}{2\sqrt{c^2+16}} \] \[ 13c^2 = 4c^2 + 64 \] \[ 9c^2 = 64 \] \[ c = \frac{8}{3} \] \[ c \approx 2.6 \] \[ \boxed{c = 2.6} \]