Question

the absolute refraction index of water and kerosene are 1.33 and 1.44 respectively.

(1) in which of the above mediums is speed of light minimum?

(2) calculate the refractive index of kerosene with respect to water?​

Answer 1

Explanation:

Given:-

• Absolute refractive index of water ,nw = 1.33
• Absolute refractive index of kerosene ,nk = 1.44

To Find:-

• (1) In which medium the speed of light is minimum .
• (2) Refractive index of kerosene with respect to water

Solution:-

As we know that absolute refractive index is defined as the speed of light in vaccum divided by speed of light in given medium .

• If refractive index of medium is greater, then the speed of light in that medium is lesser.
• If the r refractive index of medium is less, then the speed of light in that medium is greater .

(1) Here the refractive index of water is lower than refractive index of kerosene so the speed of light in kerosene is minimum .

Now, calculating the refractive index of kerosene with respect to water .

• Refractive index of kerosene with respect to water = ng/nk

Substitute the value we get

→ Refractive index of kerosene with respect to water = 1.44/1.33

→ Refractive index of kerosene with respect to water = 1.08

• Hence, the refractive index of kerosene with respect to water is 1.08

Answer 2

Answer:

Answers :-

1.

Here, the given refractive index are

$\sf \: Refractive \: index \: \begin{cases} \sf \: Water = 1.33 \\ \sf \: Kerosene = 1.44 \end{cases}$

We know that,

The things which have maximum refractive index, the things has minimum speed of light.

Here,

1.33 < 1.44

So,

Kerosene has minimum speed of light.

_____________________________

2.

Now,

We know that

$\large \sf \: R = \dfrac{ ng}{nk}$

R = Refractive index

ng = Kerosene

nk = water

R = 1.44/1.33

R = 1.08/1

R = 1.08