Question 12: Ray Optics

Question

Light ray incident along a vector $\overrightarrow{A O}(\overrightarrow{A O}=2 \hat{i}-3 \hat{j})$ emerges out along vector $\overrightarrow{O B}(\overrightarrow{O B}=C \hat{i}-4 \hat{j})$ as shown in the figure below. The value of $C$ is $\_\_\_\_$ .

Accepted Answer

A light ray travels from a medium with refractive index $\mu_1=1$ (air) to a medium with $\mu_2=1.5$. The incident vector is $\overrightarrow{\mathrm{AO}}=2 \hat{\imath}-3 \hat{\jmath}$ and the refracted vector is $\overrightarrow{\mathrm{OB}}=C \hat{\imath}-4 \hat{\jmath}$.
To use Snell's Law, we need the angles of incidence and refraction relative to the normal.
The incident ray is along the vector $\overrightarrow{\mathrm{AO}}$ points towards the origin.
$$ 	an 	heta_1=\frac{	ext { Horizontal component }}{	ext { Vertical component }}=\frac{2}{3} $$
So, using the trigonometric identity,
$$ \sin 	heta_1=\frac{2}{\sqrt{2^2+3^2}}=\frac{2}{\sqrt{13}} $$
The refracted ray is along the vector $\overrightarrow{\mathrm{OB}}$ points away from the origin. The angle with the vertical is:
$$ 	an 	heta_2=\frac{C}{4} $$
From this, $\sin 	heta_2$ is:
$$ \sin 	heta_2=\frac{C}{\sqrt{C^2+4^2}}=\frac{C}{\sqrt{C^2+16}} $$
Applying Snell's Law
$$ \mu_1 \sin 	heta_1=\mu_2 \sin 	heta_2 $$
Substituting the values :
$$ 1 	imes\left(\frac{2}{\sqrt{13}}ight)=1.5 	imes\left(\frac{C}{\sqrt{C^2+16}}ight) $$
$$\Rightarrow $$ $\frac{2}{\sqrt{13}}=\frac{3}{2} 	imes \frac{C}{\sqrt{C^2+16}}$
$$\Rightarrow $$ $$ 4 \sqrt{C^2+16}=3 C \sqrt{13} $$
Squaring both sides,
$$ 16\left(C^2+16ight)=9 C^2(13) $$
$$\Rightarrow $$ $16 C^2+256=117 C^2$
$$\Rightarrow $$ $256=117 C^2-16 C^2$
$$\Rightarrow $$ $256=101 C^2$
$$\Rightarrow $$ $C^2=\frac{256}{101} \approx 2.53$
$$\Rightarrow $$ $$ C=\sqrt{2.53} \approx 1.6 $$
Therefore, the value of C is 1.6 .
Hence, the correct option is (A).

Answer

1.6

Answer

0.16

Answer

11.6

Answer

16

Ray Optics

Light ray incident along a vector \overrightarrow{A O}(\overrightarrow{A O}=2 \hat{i}-3 \hat{j}) emerges out along vector \overrightarrow{O B}(\overrightarrow{O B}=C \hat{i}-4 \hat{j}) as shown in the figure below. The value of C is \_\_\_\_ .

Choose the correct answer: