Understanding Refractive Index and Velocity of Light

The question explores the relationship between the refractive indices of two media and the speeds at which light travels through them. This relationship is a fundamental concept in optics.

Key Concept: Refractive Index

The refractive index of a medium is a measure of how much the speed of light is reduced when it passes through that medium compared to its speed in a vacuum. The absolute refractive index ($n$) of a medium is defined by the formula:

$$ n = \frac{c}{v} $$

Where:

• $c$ is the speed of light in a vacuum (approximately $3 \times 10^8$ m/s).
• $v$ is the speed of light in the medium.

A higher refractive index means light travels slower in that medium.

Relating Refractive Indices and Velocities in Two Media

Consider two media, Medium 1 and Medium 2, with refractive indices $n_1$ and $n_2$, and velocities of light $v_1$ and $v_2$, respectively.

For Medium 1: $$ n_1 = \frac{c}{v_1} $$

For Medium 2: $$ n_2 = \frac{c}{v_2} $$

If we consider the ratio of the refractive indices, we get:

$$ \frac{n_2}{n_1} = \frac{c/v_2}{c/v_1} = \frac{c}{v_2} \times \frac{v_1}{c} = \frac{v_1}{v_2} $$

This gives us a crucial relationship: The ratio of the refractive indices of two media is equal to the inverse ratio of the speeds of light in those media. $$ \frac{n_2}{n_1} = \frac{v_1}{v_2} $$

Applying the Given Information

The question states that the ratio of the refractive indices $n_2/n_1$ is 1.5.

So, we have:

$$ \frac{n_2}{n_1} = 1.5 $$

Using the relationship derived above, we can write:

$$ \frac{v_1}{v_2} = 1.5 $$

To find the relationship between $v_1$ and $v_2$, we can rearrange this equation:

$$ v_1 = 1.5 \times v_2 $$

This equation tells us that the velocity of light in Medium 1 ($v_1$) is 1.5 times the velocity of light in Medium 2 ($v_2$).

Analyzing the Options

Let's examine the given options in light of our finding $v_1 = 1.5 v_2$:

• Option 1: v₁ is 1.5 times v₂. This statement matches our derived relationship $v_1 = 1.5 v_2$.
• Option 2: v₂ is 1.5 times v₁. This means $v_2 = 1.5 v_1$. If we rearrange this, we get $v_1/v_2 = 1/1.5 = 1/(3/2) = 2/3$. This contradicts the given information $n_2/n_1 = v_1/v_2 = 1.5$.
• Option 3: v₁ is equal to v₂. This means $v_1/v_2 = 1$. This contradicts the given information $v_1/v_2 = 1.5$.
• Option 4: v₂ is equal to v₁. This is the same as Option 3 and is incorrect.

Based on the analysis, the statement that $v_1$ is 1.5 times $v_2$ is the correct one.

Conclusion on Refractive Index and Velocity

When the refractive index of medium 2 relative to medium 1 ($n_2/n_1$) is greater than 1 (in this case, 1.5), it implies that medium 2 is optically denser than medium 1. Light travels slower in an optically denser medium. Our result shows that $v_1 = 1.5 v_2$, meaning $v_1 > v_2$. This is consistent: if $n_2 > n_1$ (because $n_2/n_1 = 1.5 > 1$), then $v_2 < v_1$. Specifically, $v_1$ is indeed greater than $v_2$, confirming our derived relationship.

Revision Table: Refractive Index Concepts

Additional Information: Optical Density

The concept of refractive index is closely related to optical density. Optical density is not the same as physical density (mass per unit volume), although denser materials often have higher optical density. A medium with a higher refractive index is said to be optically denser than a medium with a lower refractive index. Light travels slower in an optically denser medium.

• If $n_2 > n_1$, Medium 2 is optically denser than Medium 1, and $v_2 < v_1$.
• If $n_2 < n_1$, Medium 2 is optically rarer than Medium 1, and $v_2 > v_1$.
• In this problem, $n_2/n_1 = 1.5$, which means $n_2 = 1.5 n_1$. Since $1.5 > 1$, $n_2 > n_1$. Thus, Medium 2 is optically denser than Medium 1, confirming that $v_2$ must be less than $v_1$. Our result $v_1 = 1.5 v_2$ shows $v_1 > v_2$, which is consistent.