An ice cube (n = 1.31) has a layer of water (n = 1.33) on it. What is the critical angle for a ray in water trying to enter the ice?

(Unit=deg)

Explanation:

Given that,

The Refractive index of glass is

n(ice) = 1.31

The refractive index of water is

n(water) = 1.33

What is the critical angle?

From water to ice

Critical angle is given as.

Using Snell's Rule.

Sin( θc ) = n1 / n2

for total internal reflection the incident ray is always in the denser medium! So, water is the more denser than ice.

Then, .

Sin( θc ) = n(ice) / n(water)

Sin( θc ) = 1.31 / 1.33

Sin( θc ) = 0.98496

θc = arcSin(0.98496)

θc = 80.05°

The critical angle is 80.05°

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490 n

f=125n a=2.5m/s2