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Inga [223]
3 years ago
12

If the speed of light in air is 3.00 · 108 m/s (approximately equal to that in a vacuum) and the speed of light in water is 2.25

· 108 m/s, what is the index of refraction of water?
Physics
2 answers:
podryga [215]3 years ago
7 0

Answer: 1.333

Explanation:

Index of refraction= n=c/v

Where c is the speed of light in vacuum and v is the speed of light in material ( water) in this case.

n = 3.00.108/2.25.108

n =3.00/2.25

n = 1.333

Approximately 1.33

There is no unit because m/s cancels out m/s

Natasha_Volkova [10]3 years ago
3 0
N= velocity in vacuum/velocity in water 
n=3.00*10⁸m/s/2.25 *10⁸m/s= 3.00/2.25 ≈ 1.33

n ≈ 1.33
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Answer:

T_b=107.3784\ ^{\circ}C

Explanation:

Given:

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<u>From the Fourier's law of conduction we have:</u>

\dot Q=k.A.\frac{dT}{dx}

\dot Q=k\times \pi.r^2\times \frac{T_b-T_t}{5.2\times 10^{-3}}

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A= area of the surface through which conduction occurs

T_b= temperature of the bottom surface

15405.67=240\times \pi\times 0.12^2\times \frac{T_b-100}{5.2\times 10^{-3}}

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Explanation:

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