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**Hint:**The velocity of light through any medium will be less than the velocity of light in vacuum. The ratio of sine of angle of incidence $\left( i \right)$ to the sine of angle of refraction $\left( r \right)$ is a constant and is called refractive index. The absolute refractive index can also be expressed as the ratio of the velocity of light in vacuum $\left( c \right)$ to the velocity of light $\left( v \right)$ in a medium.

**Formula used:**

Refractive index, $\mu = \dfrac{c}{v}$

Where $\mu $ stands for the refractive index of the medium, $c$ stands for the velocity of light in vacuum, and $v$ stands for the velocity of light in a medium.

**Complete step by step solution:**

The absolute refractive index is the ratio of the velocity of light in vacuum to the velocity of light through a medium, that is

$\mu = \dfrac{c}{v}$………………………………………(A)

The velocity of light through vacuum, $c = 3 \times {10^8}$

Here we have to find the speed of light when it travels through water.

The refractive index of water is given by,

${\mu _{water}} = \dfrac{4}{3}$

If $v$ is the velocity of the light through water, then equation (A) can be rearranged to get $v$

$v = \dfrac{c}{\mu }$

$v = \dfrac{{3 \times {{10}^8}}}{{\dfrac{4}{3}}} = \dfrac{{3 \times {{10}^8} \times 3}}{4} $

$v = 2.25 \times {10^8} $

**Therefore, the answer is: Option (B), The speed of light in water is $2.25 \times {10^8}m{s^{ - 1}}.$**

**Note:**Alternate method:

The velocity of light can be calculated using another method

Since the light enters into water from air there will be a change in refractive index.

Let ${n_1}$ be the refractive of the first medium and ${n_2}$ be the refractive index of the second medium.

Let ${v_1}$ be the velocity of light through the first medium and ${v_2}$ be the velocity through the second medium.

The ratio of the velocity through second medium to the velocity through first medium will be equal to the ratio of refractive index of the first medium to the refractive index of the second medium, that is,

$\dfrac{{{v_2}}}{{{v_1}}} = \dfrac{{{n_1}}}{{{n_2}}}$………………..(B)

We know that the first medium is air and the second medium is water,

\[ \therefore {v_1} = c = 3 \times {10^8}m{s^{ - 1}} \]

$ {n_1} = 1$

${n_2} = \dfrac{4}{3} = 1.33 $

(The refractive index of air is $1$ and the refractive index of water is given)

Rearranging equation (B)

${v_2} = {v_1}\dfrac{{{n_1}}}{{{n_2}}} = 3 \times {10^8} \times \dfrac{1}{{1.33}} = 2.25 \times {10^8}m{s^{ - 1}}$

We will get the same answer as the first method.

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