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Hint: Snell’s law is used to relate between angle of incidence and angle of reflection with the refractive index of a medium when light travels from one medium to another medium.

When light travels from one medium to another medium it changes its direction or velocity or both. This is called refraction of light.

Snell’s law of refraction of light: This is coined by “Willebrand Snell’ and this is the second law of refraction of light also called as “law of sines”.

Statement: It states that “The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media and given wavelength of light.”

The constant in law is called refractive index and denoted by the letter ’n’.

When a ray of light travels from medium 1 to medium 2 then it is the refractive index of medium 2 w.r.t medium1 and is written as ${{n}_{21}}$. For example, if a ray of light is travelling from water to glass, it is the refractive index of glass w.r.t water.

This ${{n}_{21}}$is called relative refractive index and is given by,

${{n}_{21}}=\dfrac{{{n}_{2}}}{{{n}_{1}}}$

Now mathematical form of Snell’s law is given by,

${{n}_{21}}=\dfrac{\sin i}{\sin r}$

Where, i - is the angle of incidence

r- is the angle of refraction

$\begin{align}

& \dfrac{{{n}_{2}}}{{{n}_{1}}}=\dfrac{\sin i}{\sin r} \\

& {{n}_{2}}\sin r={{n}_{1}}\sin i \\

\end{align}$

$\Rightarrow {{n}_{1}}\sin i={{n}_{2}}\sin r$

This is Snell’s law.

Now for the second question:

Higher the refractive index means the medium is denser. Let us consider Snell’s law,

$\Rightarrow {{n}_{1}}\sin i={{n}_{2}}\sin r$

When light travels from denser to rarer medium

$\begin{align}

& {{\text{n}}_{1}}\text{ } > \text{ }{{\text{n}}_{2}} \\

& \Rightarrow \dfrac{{{\text{n}}_{1}}}{{{\text{n}}_{2}}} > 1 \\

& \Rightarrow \dfrac{\text{sinr}}{\text{sini}} > 1 \\

\end{align}$

For an angle less than 90°, sine of an angle increases with angle, so

$\text{ r } > \text{ i}$

Which means angle of refraction is greater than angle of incidence. So, light bends away from the normal making $\text{ r } > \text{ i}$.

Similarly, when light travels from rarer to denser medium-

$\text{ r } < \text{ i}$

Which means the angle of refraction is smaller than angle of incidence. So, light bends towards normal making $\text{ r } < \text{ i}$.

In the above question, the refracted ray bends away from the normal which means that it travelled from denser medium A to rarer medium B.

Therefore, A is optically denser than B.

Note: While writing Snell’s law student may get confused and write angle of reflection instead of angle of refraction which is incorrect Snell’s law is for refraction of light.

When light travels from one medium to another medium its velocity and wavelength changes as both are inversely proportional to the refractive index of a medium.

i.e., $\dfrac{{{n}_{2}}}{{{n}_{1}}}=\dfrac{{{v}_{1}}}{{{v}_{2}}}=\dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}$

where frequency does not depend on the medium but depends on the source of light.

But change in velocity or wavelength are not the reason for the bending of light.

When light travels from one medium to another medium it changes its direction or velocity or both. This is called refraction of light.

__Complete step by step answer:__Snell’s law of refraction of light: This is coined by “Willebrand Snell’ and this is the second law of refraction of light also called as “law of sines”.

Statement: It states that “The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media and given wavelength of light.”

The constant in law is called refractive index and denoted by the letter ’n’.

When a ray of light travels from medium 1 to medium 2 then it is the refractive index of medium 2 w.r.t medium1 and is written as ${{n}_{21}}$. For example, if a ray of light is travelling from water to glass, it is the refractive index of glass w.r.t water.

This ${{n}_{21}}$is called relative refractive index and is given by,

${{n}_{21}}=\dfrac{{{n}_{2}}}{{{n}_{1}}}$

Now mathematical form of Snell’s law is given by,

${{n}_{21}}=\dfrac{\sin i}{\sin r}$

Where, i - is the angle of incidence

r- is the angle of refraction

$\begin{align}

& \dfrac{{{n}_{2}}}{{{n}_{1}}}=\dfrac{\sin i}{\sin r} \\

& {{n}_{2}}\sin r={{n}_{1}}\sin i \\

\end{align}$

$\Rightarrow {{n}_{1}}\sin i={{n}_{2}}\sin r$

This is Snell’s law.

Now for the second question:

Higher the refractive index means the medium is denser. Let us consider Snell’s law,

$\Rightarrow {{n}_{1}}\sin i={{n}_{2}}\sin r$

When light travels from denser to rarer medium

$\begin{align}

& {{\text{n}}_{1}}\text{ } > \text{ }{{\text{n}}_{2}} \\

& \Rightarrow \dfrac{{{\text{n}}_{1}}}{{{\text{n}}_{2}}} > 1 \\

& \Rightarrow \dfrac{\text{sinr}}{\text{sini}} > 1 \\

\end{align}$

For an angle less than 90°, sine of an angle increases with angle, so

$\text{ r } > \text{ i}$

Which means angle of refraction is greater than angle of incidence. So, light bends away from the normal making $\text{ r } > \text{ i}$.

Similarly, when light travels from rarer to denser medium-

$\text{ r } < \text{ i}$

Which means the angle of refraction is smaller than angle of incidence. So, light bends towards normal making $\text{ r } < \text{ i}$.

In the above question, the refracted ray bends away from the normal which means that it travelled from denser medium A to rarer medium B.

Therefore, A is optically denser than B.

Note: While writing Snell’s law student may get confused and write angle of reflection instead of angle of refraction which is incorrect Snell’s law is for refraction of light.

When light travels from one medium to another medium its velocity and wavelength changes as both are inversely proportional to the refractive index of a medium.

i.e., $\dfrac{{{n}_{2}}}{{{n}_{1}}}=\dfrac{{{v}_{1}}}{{{v}_{2}}}=\dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}$

where frequency does not depend on the medium but depends on the source of light.

But change in velocity or wavelength are not the reason for the bending of light.

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