This problem aims to find the refractive index of a prism having angles of $30\space60$ and $90$ degrees. The concepts required to solve this problem are related to snell’s law and the index of refraction. Now the refractive index is defined as the ratio of the speed of light in any medium (e.g. water), to the speed of light in a vacuum.
The Refractive index is also known as the refraction index or the index of refraction. Whenever the light passes through a medium, its behavior tends to be different which depends on the properties of the medium.
Since the refractive index is the ratio of two quantities, so it is unitless and dimensionless. It is a numerical value that demonstrates how slow the light would be in the material than it is in the vacuum by displaying a number. The refractive index is denoted by the symbol $\eta$, which is the ratio of the velocity of light in a vacuum and the velocity of light in a medium. The formula to find the refractive index is shown as:
\[ \eta = \dfrac{c}{v} \]
Where,
$\eta$ is the refractive index,
$c$ is the speed of light in a vacuum that is $3\times 10^8\space m/s$,
$v$ is the speed of light in any substance.
Expert Answer
To solve this problem, we must be familiar with Snell’s Law, which is similar to the refractive index formula:
\[ \dfrac{\sin \phi}{\sin \theta} = \dfrac{n_1}{n_2} = constant = \eta \]
Where,
$\theta$ is the angle of incidence, and $\phi$ is the angle of refraction, $n_1$ and $n_2$ are the different mediums, and we know the $\eta$ is the refractive index.
Here, the angle of incidence $\theta$ is $30^{\circ}$ and the angle between the refracted ray and the horizontal $\theta_1$is $19.6^{\circ}$.
Now the angle of refraction $\phi$ can be calculated as:
\[\phi = \theta + \theta_1\]
Plugging in the values:
\[\phi = 30^{\circ} + 19.6^{\circ}\]
\[\phi = 49.6^{\circ}\]
Hence, we can use the angle of refraction in Snell’s Law to find the index of refraction:
\[\dfrac{\sin \phi}{\sin \theta} = \dfrac{n_1}{n_2} \]
\[\dfrac{\sin \phi}{\sin \theta}\times n_2 = n_1 \]
\[n_1 = \dfrac{\sin \phi}{\sin \theta}\times n_2 \]
Substituting the values in the above equation:
\[n_1 = \dfrac{\sin 49.6^{\circ}}{\sin 30^{\circ}}\times (1.0)\]
\[n_1 = \dfrac{0.761}{0.5}\]
\[ n_1 = 1.52\]
Numerical Result
The refractive index of the prism comes out to be $ n_1 = 1.52$.
Example
Find the refractive index of a medium in which light passes at a speed of $1.5\times 10^8 m/s$. Let’s say the refractive index of water is $\dfrac{4}{3}$ and that of acrylic is $\dfrac{3}{2}$. Find the refractive index of acrylic w.r.t. water.
The formula to find the refractive index is:
\[\eta = \dfrac{c}{v} \]
Substituting the values in the equation, we get
\[\eta = \dfrac{3 \times 10^8 m/s}{1.5\times 10^8 m/s} = 2\]
The refractive index comes out to be $2$.
Now $\eta_w = \dfrac{4}{3}$ and $\eta_a = \dfrac{3}{2}$
The Refractive index of acrylic w.r.t. water is:
\[\eta^{w}_{a} = \dfrac{\eta_a}{\eta_w} \]
\[= \dfrac{\dfrac{3}{2}}{\dfrac{4}{3}} \]
\[= {\dfrac{9}{8}}\]