Question

State Snell’s law of refraction of light. Write an expression to relate refractive index of a medium with speed of light in vacuum.The refractive index of a medium ‘a’ with respect to medium ‘b’ is 2/3 and the refractive index of medium ‘b’ with respect to medium ‘c’ is 4/3. Find the refractive index of medium ‘c’ with respect to medium ‘a’. (5)

Answer 1

Explanation:

Snell's Law of refraction : The ratio of sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given colour and for the given pair of media. Hence, refractive index of medium 'c' with respect to medium 'a' is 9/8

Answer 2

Answer:

Snell's Law:

• The relation between the angle of incidence and the angle of refraction was discovered by a Dutch physicist Willebrord Snell in 1621 and is known as Snell's law.
• This law forms an important part of the laws of refraction which are stated as follows:

1. The incident ray and the refracted ray are on the opposite sides of the normal at the point of incidence and all three lie in the same plane.
2. For a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant.

• The constant is called the refractive index of the second medium with respect to the first medium.

• It is written as $n^1_2 = \frac{sin(i)}{sin(r)} = \frac{n_2}{n_1}$ , where $n_1$ is the refractive index of the first medium and $n_2$ is the refractive index of the second medium

The refractive index of a medium with the speed of light in a vacuum is written as $n=\frac{c}{v}$, where n is the refractive index of the medium, and c is the speed of light in a vacuum.

The refractive index of medium 'c' with respect to medium 'a' is 9/8.

Explanation:

Given:

Refractive index of medium 'a' with respect to medium 'b' ($n_a/n_b$) = 2/3

Refractive index of medium 'b' with respect to medium 'c' ($n_b/n_c$) = 4/3

To find:

Refractive index of medium 'c' with respect to medium 'a' ($n_c/n_a$)

Solution:

$\frac{n_a}{n_c} = \frac{n_a}{n_b}*\frac{n_b}{n_c}$

Substituting the given values, we get

$\frac{n_a}{n_c} = \frac{2}{3}*\frac{4}{3}$

$\frac{n_a}{n_c} = \frac{8}{9}$

$\frac{n_c}{n_a} = \frac{1}{\frac{n_a}{n_c} }$

$\frac{n_c}{n_a} = \frac{1}{\frac{8}{9} }$

$\frac{n_c}{n_a} = \frac{9}{8} }$

Therefore, the refractive index of medium ‘c’ with respect to medium ‘a’ is 9/8.

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