Light of wavelength 550 nm in air is found to travel at 1.96 x 10^8 m/s in a certain liquid. Determine the index of refraction of the liquid.

To determine the index of refraction of the liquid, we can use Snell's law, which relates the angles of incidence and refraction of a ray of light passing through a boundary between two transparent media with different refractive indices.

Snell's law states:

n₁ * sin(θ₁) = n₂ * sin(θ₂)

Where:
- n₁ and n₂ are the indices of refraction of the first and second media, respectively
- θ₁ and θ₂ are the angles of incidence and refraction, respectively

In this case, we are given the wavelength of light in air (550 nm) and the speed of light in the liquid (1.96 × 10^8 m/s). We need to find the index of refraction of the liquid.

To begin, we need to convert the wavelength from nanometers to meters. Since 1 meter is equal to 1 × 10^9 nanometers, the wavelength in meters would be:

λ = 550 nm = 550 × 10^-9 m

Next, we can use the formula for the speed of light in a medium:

c = λ * f

Where:
- c is the speed of light in a medium
- λ is the wavelength of light
- f is the frequency of light

We know the speed of light in the liquid (c₂ = 1.96 × 10^8 m/s) and the wavelength in the liquid (λ₂), but we need to find the frequency in the liquid.

Since the frequency of light remains constant when it passes from one medium to another, we can say that:

f₁ = f₂

Therefore:

c₁/λ₁ = c₂/λ₂

We can now rearrange this equation to solve for the wavelength in the liquid (λ₂):

λ₂ = (c₂ * λ₁) / c₁

Plugging in the given values and solving for λ₂:

λ₂ = (1.96 × 10^8 m/s * 550 × 10^-9 m) / 3 × 10^8 m/s

λ₂ = 0.3582

Now, we can use the relationship between the wavelength and the speed of light in a medium to find the index of refraction of the liquid:

n₂ = c₁ / c₂

Plugging in the values:

n₂ = (3 × 10^8 m/s) / (0.3582)

n₂ = 836,360.2

Therefore, the index of refraction of the liquid is approximately 836,360.2.