Suppose you have an unknown clear substance immersed in water, and you wish to identify it by finding its index of refraction. You arrange to have a beam of light enter it at an angle of 45.7∘, and you observe the angle of refraction to be 30.5∘. What is the index of refraction of the substance? Water has an index of refraction equal to 1.333.

Question

Jethro

Physics

7 January, 00:54

Suppose you have an unknown clear substance immersed in water, and you wish to identify it by finding its index of refraction. You arrange to have a beam of light enter it at an angle of 45.7∘, and you observe the angle of refraction to be 30.5∘. What is the index of refraction of the substance? Water has an index of refraction equal to 1.333.

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Reece Robles · Answer 1

Answer: 1.65

Explanation: Snell's law says that index of refraction of the first material times sine theta one equals index of refraction of the second material, which we don't know, multiplied by sine theta two,. So, we'll solve this for N two by dividing both sides by sine theta two. So, N two is N one sine theta one over sine theta two. So, that's 1.33 index of refraction of water times sine of 45.7degrees incidence angle divided by sine of 30.5 degrees angle of refraction, giving us 1.65.

The formula for snell's law is N1*sintheta1=N2*sinetheta2