
If the difference in velocities of light in glass and water is $0.25\times {{10}^{8}}\,m/s$. Find the velocity of light in the air. (Refractive index of water is 4/3 and refractive index of glass is 3/2)
Answer
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Hint: The refractive index of a medium is defined as the ratio of velocity of wave in the medium to velocity of wave in vacuum or air. Use this definition to derive relations between velocity of light in different media. Then solve these equations for velocity of light in air.
Complete answer:
Let us assume that velocity of light in air is $c$.
Let velocity of light in water and glass be ${{v}_{w}}$ and ${{v}_{g}}$ respectively.
The refractive index of a medium is defined as the ratio of velocity of wave in the medium to velocity of wave in vacuum or air.
We are given that the refractive index of water is 4/3. According to the definition of refractive index.
${{\mu }_{water}}=\dfrac{4}{3}=\dfrac{c}{{{v}_{w}}}$
On cross multiplying, we get
$\dfrac{4}{3}{{v}_{w}}=c$ …… (1)
Similarly, from refractive index of glass, we can obtain
$\dfrac{3}{2}{{v}_{g}}=c$ …… (2)
Taking ratio of (1) and (2), we get
$\dfrac{\dfrac{4}{3}{{v}_{w}}}{\dfrac{3}{2}{{v}_{g}}}=1\Rightarrow 8{{v}_{w}}=9{{v}_{g}}$
This implies that, $8{{v}_{w}}-9{{v}_{g}}=0$ …… (3)
The difference in velocities of light in glass and water is $0.25\times {{10}^{8}}\,m/s$. This can be written as,
${{v}_{w}}-{{v}_{g}}=0.25\times {{10}^{8}}\,m/s$ …… (4)
Solving (3) and (4) for velocity of light in glass by elimination method, we get
$8\left( {{v}_{w}}-{{v}_{g}} \right)-\left( 8{{v}_{w}}-9{{v}_{g}} \right)={{v}_{g}}=8\times 0.25\times {{10}^{8}}m/s$
Simplifying this equation, we have
${{v}_{g}}=2\times {{10}^{8}}m/s$
We substitute this value in equation (2),
$\dfrac{3}{2}\times 2\times {{10}^{8}}m/s=c$
This implies that velocity of light in air is $c=3\times {{10}^{8}}\,m/s$
Note:
The refractive index of a medium is defined as the ratio of velocity of wave in the medium to velocity of wave in vacuum or air. It is also defined as the ratio of the sines of the angles of incidence and refraction where incidence is in vacuum.
Complete answer:
Let us assume that velocity of light in air is $c$.
Let velocity of light in water and glass be ${{v}_{w}}$ and ${{v}_{g}}$ respectively.
The refractive index of a medium is defined as the ratio of velocity of wave in the medium to velocity of wave in vacuum or air.
We are given that the refractive index of water is 4/3. According to the definition of refractive index.
${{\mu }_{water}}=\dfrac{4}{3}=\dfrac{c}{{{v}_{w}}}$
On cross multiplying, we get
$\dfrac{4}{3}{{v}_{w}}=c$ …… (1)
Similarly, from refractive index of glass, we can obtain
$\dfrac{3}{2}{{v}_{g}}=c$ …… (2)
Taking ratio of (1) and (2), we get
$\dfrac{\dfrac{4}{3}{{v}_{w}}}{\dfrac{3}{2}{{v}_{g}}}=1\Rightarrow 8{{v}_{w}}=9{{v}_{g}}$
This implies that, $8{{v}_{w}}-9{{v}_{g}}=0$ …… (3)
The difference in velocities of light in glass and water is $0.25\times {{10}^{8}}\,m/s$. This can be written as,
${{v}_{w}}-{{v}_{g}}=0.25\times {{10}^{8}}\,m/s$ …… (4)
Solving (3) and (4) for velocity of light in glass by elimination method, we get
$8\left( {{v}_{w}}-{{v}_{g}} \right)-\left( 8{{v}_{w}}-9{{v}_{g}} \right)={{v}_{g}}=8\times 0.25\times {{10}^{8}}m/s$
Simplifying this equation, we have
${{v}_{g}}=2\times {{10}^{8}}m/s$
We substitute this value in equation (2),
$\dfrac{3}{2}\times 2\times {{10}^{8}}m/s=c$
This implies that velocity of light in air is $c=3\times {{10}^{8}}\,m/s$
Note:
The refractive index of a medium is defined as the ratio of velocity of wave in the medium to velocity of wave in vacuum or air. It is also defined as the ratio of the sines of the angles of incidence and refraction where incidence is in vacuum.
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