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 **refrac**t**ive 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 **S****nell’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:

\[\eta^{w}_{a} = \dfrac{\eta_a}{\eta_w} \]

\[= \dfrac{\dfrac{3}{2}}{\dfrac{4}{3}} \]

\[= {\dfrac{9}{8}}\]