Question

A ray of light moving along the unit vector $\left( { - i - 2j} \right)$ undergoes refraction at an interface of two media, which is the x-z plane. The refractive index for y > 0 is 2 while for y < 0, it is $\sqrt 5 /2$. The unit vector along which the refracted ray moves is:
A. $\dfrac{{\left( { - 3i - 5j} \right)}}{{\sqrt {34} }}$
B. $\dfrac{{\left( { - 4i - 3j} \right)}}{5}$
C. $\dfrac{{\left( { - 3i - 4j} \right)}}{5}$
D. None of these

Answer 1

The above problem can be resolved using the fundamentals of the refractive index, along with the fundamentals of the angle of incidence and angle of refraction when a ray of light strikes on a plane reflected surface, such that angles of incidence and refraction are formed. Moreover, these angles are related to another component that determines the bending of light rays in a medium; this component is known as the refractive index.

Complete step by step answer:
The vector representation of the incident ray is, $\vec r = \left( { - i - 2j} \right)$.
The refractive index for y > 0 is, ${\mu _1} = 2$.
The refractive index for y < 0 is, ${\mu _2} = \sqrt 5 /2$
Let r be the refracted angle. Then, the value of r in the vector form is given as,
$\vec r\left( { - j} \right) = \left| r \right|\cos i$
Here, i is the incident angle.
Solve by substituting the values as,

\vec r\left( { - j} \right) = \left| r \right|\cos i\\\ \left( { - i - 2j} \right)\left( { - j} \right) = \left( {\sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 2} \right)}^2}} } \right)\cos i\\\ \left( { - i - 2j} \right)\left( { - j} \right) = \sqrt 5 \cos i \end{array}$$$$$$ Taking the values from the above equation as, $$\begin{array}{l} \left( { - i - 2j} \right)\left( { - j} \right) = \sqrt 5 \cos i\\\ \cos i = \dfrac{2}{{\sqrt 5 }}\\\ \sin i = \dfrac{1}{{\sqrt 5 }} \end{array}$$ Now, apply the Snell’s law as, $$\begin{array}{l} {\mu _1}\sin i = {\mu _2}\sin r\\\ 2 \times \left( {\dfrac{1}{{\sqrt 5 }}} \right) = \dfrac{{\sqrt 5 }}{2}\sin r\\\ \sin r = \dfrac{4}{5} \end{array}$$ So, the value of cosine of refraction angle is, $$\cos r = \dfrac{3}{5}$$ The vector form of the refraction angle is, $$\begin{array}{l} \vec r\left( { - j} \right) = \left| r \right|\cos r\\\ \left( { - i - yj} \right)\left( { - j} \right) = \left( {\sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - y} \right)}^2}} } \right)\cos r\\\ \left( { - i - yj} \right)\left( { - j} \right) = \sqrt {1 + {y^2}} \cos r \end{array}$$ Here, y is any constant. $$$$ Further solving as, $$\begin{array}{l} \left( { - i - yj} \right)\left( { - j} \right) = \sqrt {1 + {y^2}} \cos r\\\ \left( { - i - yj} \right)\left( { - j} \right) = \sqrt {1 + {y^2}} \times \left( {\dfrac{3}{5}} \right)\\\ 5y = 3\sqrt {1 + {y^2}} \\\ y = \dfrac{3}{4} \end{array}$$ $$$$ The unit vector along which the refracted ray is given as, $$\begin{array}{l} {{\vec r}_c} = \dfrac{{\left( { - \hat i - y\hat j} \right)}}{{\left| {{r_c}} \right|}}\\\ {{\vec r}_c} = \dfrac{{\left( { - \hat i - y\hat j} \right)}}{{\sqrt {1 + {y^2}} }}\\\ {{\vec r}_c} = \dfrac{{\left( { - \hat i - \left( {\dfrac{3}{4}} \right)\hat j} \right)}}{{\sqrt {1 + {{\left( {\dfrac{3}{4}} \right)}^2}} }}\\\ {{\vec r}_c} = \dfrac{{ - 4\hat i - 3\hat j}}{5} \end{array}$$ Therefore, the unit vector along which the refracted ray is $$\dfrac{{ - 4\hat i - 3\hat j}}{5}$$ and option (B) is correct. **Note:** To solve the given problem, one needs to understand the concept behind the refraction and reflection. When the light ray is being refracted through the surface, the angle so formed is called the angle of refraction and similarly, the rays where reflection of light rays takes place, the angle corresponding to it is known as the angle of incidence.