The refractive index of kerosene, turpentine and water are 1.44, 1.47 and 1.33 respectively, then
A. the speed of light is maximum and minimum in kerosene and turpentine
B. the speed of light is maximum and minimum in water and turpentine
C. the speed of light is maximum and minimum in turpentine and water
D. the speed of light is maximum and minimum in turpentine and kerosene
Answer
601.2k+ views
Hint: Refractive index or the index of refraction gives the measure of bending of light ray when it moves from one medium to another. Express the refractive index in terms of the speed of light in vacuum and the speed of light in the medium. Then put the given values to see in which medium the speed is maximum and in which medium the velocity is minimum.
Complete step-by-step answer:
The refractive index of a medium can be defined as the ratio of the sine of the angle of incidence and the angle of refraction of light when light passes through the medium.
We can also define the refractive index of a medium in terms of the velocity of light in the medium and the velocity of light in vacuum. Refractive index of a medium can be expressed as the ratio of the speed of light in vacuum and speed of light in the medium.
Mathematically we can write,
$n=\dfrac{c}{v}$
Where n is the refractive index of the medium, c is the speed of light in vacuum and v is the speed of light in the medium.
So, we can write,
$v=\dfrac{c}{n}$
Now, the refractive index of kerosene, turpentine and water are 1.44, 1.47 and 1.33 respectively.
So, the speed of light in kerosene will be, ${{v}_{k}}=\dfrac{c}{1.44}$
So, the speed of light in turpentine will be, ${{v}_{t}}=\dfrac{c}{1.47}$
So, the speed of light in water will be, ${{v}_{w}}=\dfrac{c}{1.33}$
From the above, we can say that the speed of light is maximum in water and minimum in turpentine.
So, the correct answer is “Option B”.
Note: When light ray enters another medium from one medium, the direction of the light ray changes. Here, the refractive index of the medium can be given in terms of the angle of incidence and the angle of refraction. $n=\dfrac{\sin i}{\sin r}$
Complete step-by-step answer:
The refractive index of a medium can be defined as the ratio of the sine of the angle of incidence and the angle of refraction of light when light passes through the medium.
We can also define the refractive index of a medium in terms of the velocity of light in the medium and the velocity of light in vacuum. Refractive index of a medium can be expressed as the ratio of the speed of light in vacuum and speed of light in the medium.
Mathematically we can write,
$n=\dfrac{c}{v}$
Where n is the refractive index of the medium, c is the speed of light in vacuum and v is the speed of light in the medium.
So, we can write,
$v=\dfrac{c}{n}$
Now, the refractive index of kerosene, turpentine and water are 1.44, 1.47 and 1.33 respectively.
So, the speed of light in kerosene will be, ${{v}_{k}}=\dfrac{c}{1.44}$
So, the speed of light in turpentine will be, ${{v}_{t}}=\dfrac{c}{1.47}$
So, the speed of light in water will be, ${{v}_{w}}=\dfrac{c}{1.33}$
From the above, we can say that the speed of light is maximum in water and minimum in turpentine.
So, the correct answer is “Option B”.
Note: When light ray enters another medium from one medium, the direction of the light ray changes. Here, the refractive index of the medium can be given in terms of the angle of incidence and the angle of refraction. $n=\dfrac{\sin i}{\sin r}$
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