To understand the relationship between the speeds of light in two different media and their refractive indices, we start by recalling Snell's Law and the definition of refractive index. The refractive index \( n \) of a medium is given by the ratio of the speed of light in vacuum \( c \) to the speed of light in that medium \( v \). Mathematically, this is represented as:
\[ n = \frac{c}{v} \]
For two different media with refractive indices \( n_1 \) and \( n_2 \) and speeds of light \( v_1 \) and \( v_2 \) respectively, the relationship between these can be expressed as:
\[ n_1 = \frac{c}{v_1} \quad \text{and} \quad n_2 = \frac{c}{v_2} \]
Dividing these two equations, we get:
\[ \frac{n_1}{n_2} = \frac{\frac{c}{v_1}}{\frac{c}{v_2}} = \frac{v_2}{v_1} \]
Rearranging the equation gives:
\[ \frac{v_1}{v_2} = \frac{n_2}{n_1} \]
Concept: The speed of light in a medium is inversely proportional to its refractive index: v ∝ 1/n
So, v1 / v2 = n2 / n1
Correct Answer: v₁/v₂ = n₂/n₁
Two slits 0.1 mm apart are arranged 1.20 m from a screen. Light of wavelength 600 nm from a distant source is incident on the slits. How far apart will adjacent bright interference fringes be on the screen?