Question

 (a) A tank containing methanol has walls 2.50 cm thick made of glass of refractive index 1.550. Light from the outside air strikes the glass at 41.3 degrees angle with the normal to the glass. Find the angle the light makes with the normal in the methanol.

(b) The tank is emptied and refilled with an unknown liquid. If light incident at the same angle as in part (a) enters the liquid in the tank at an angle of 20.2 from the normal, what is the refractive index of the unknown liquid?

Answer 1

Concepts and reason

The concept required to solve the given problem is Snell’s law of refraction.

Initially calculate the angle the light makes with the normal in the methanol and later calculate the refractive index of the unknown liquid will be calculated.

Fundamentals

Snell’s law of refraction states that the ratio of sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given color and for the given pair of media.

The expression for the Snell’s law of refraction is,

${\mu _1}\sin \left( i \right) = {\mu _2}\sin \left( r \right)$ …… (1)

Here, ${\mu _1}$ is the refractive index of air, ${\mu _2}$ is the refractive index of glass, $i$ is the angle of incidence, and $r$ is the angle of refraction.

(a)

The expression to calculate angle is,

$\;r = {\sin ^{ - 1}}\left( {\frac{{{\mu _1}\sin \left( i \right)}}{{{\mu _2}}}} \right)$

Substitute $41.3^\circ$ for i, 1 for ${\mu _1}$ and 1.55 for ${\mu _2}$ in the above equation.

$\begin{array}{c}\\r = {\sin ^{ - 1}}\left( {\frac{{\sin \left( {41.3^\circ } \right)}}{{1.55}}} \right)\\\\ = 1.55\\\end{array}$

Substitute $25.2^\circ$ for i, 1.55 for ${\mu _1}$ , and 1.329 for ${\mu _2}$ in equation (1).

$\begin{array}{c}\\r = {\sin ^{ - 1}}\left( {\frac{{1.55\sin \left( {25.2^\circ } \right)}}{{1.328}}} \right)\\\\ = 29.8^\circ \\\end{array}$

(b)

The refractive index of an unknown liquid is,

${\mu _1}\sin \left( i \right) = {\mu _2}\sin \left( r \right)$

Substitute 1.55 for ${\mu _1}$ , $25.2^\circ$ for i, and $20.2^\circ$ for r in the above equation.

$\begin{array}{c}\\{\mu _2} = \frac{{{\mu _1}\sin i}}{{\sin r}}\\\\ = \left( {\frac{{1.55\sin \left( {25.2^\circ } \right)}}{{\sin \left( {20.2^\circ } \right)}}} \right)\\\\ = 1.91\\\end{array}$

Ans: Part a

The angle made by the light ray with the normal in the methanol is $29.8^\circ$ .

Part b

The refractive index of an unknown liquid is 1.91.