Question

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

• A. 1.6
• B. 0.16
• C. 11.6
• D. 16

Answer 1

The Correct Option is D

Step 1: Find angle of incidence The interface is horizontal, so the normal is along the \(y\)-axis. For incident vector \[ \vec{v}_1=2\hat{i}-3\hat{j} \] the angle of incidence \(i\) is measured with the normal. Thus, \[ \tan i=\frac{\text{horizontal component}}{\text{vertical component}} \] \[ \tan i=\frac{2}{3} \] Hence, \[ \sin i=\frac{2}{\sqrt{2^2+3^2}} \] \[ \sin i=\frac{2}{\sqrt{13}} \] Step 2: Find angle of refraction For refracted vector \[ \vec{v}_2=C\hat{i}-4\hat{j} \] Similarly, \[ \tan r=\frac{C}{4} \] Therefore, \[ \sin r=\frac{C}{\sqrt{C^2+4^2}} \] \[ \sin r=\frac{C}{\sqrt{C^2+16}} \] Step 3: Apply Snell's law Using \[ \mu_1\sin i=\mu_2\sin r \] Given air to glass: \[ \mu_1=1,\qquad \mu_2=1.5 \] So, \[ 1\cdot \frac{2}{\sqrt{13}} = 1.5\cdot \frac{C}{\sqrt{C^2+16}} \] \[ \frac{2}{\sqrt{13}} = \frac{3C}{2\sqrt{C^2+16}} \] Squaring both sides: \[ \frac{4}{13} = \frac{9C^2}{4(C^2+16)} \] Cross-multiplying: \[ 16(C^2+16)=117C^2 \] \[ 16C^2+256=117C^2 \] \[ 101C^2=256 \] \[ C^2=\frac{256}{101} \] \[ C=\sqrt{\frac{256}{101}} \] \[ C\approx 1.59 \] \[ C\approx 1.6 \] Final Answer: \[ \boxed{1.6} \]