What is the critical angle for light traveling from crown glass (n = 1.52) into water (n = 1.33)?
The answer to the above question is that the critical angle is 61 degrees (610)
Here is the method I applied to solve the problem:
We will identify our data from the above question
The light travels from crown glass = Incident Light = ni = 1.52
It then passes through water = Refracted ray of light = nr = 1.33
According to the definition of the critical angle. When light passes through a denser to a less dense medium, it forms an angle of 900. Therefore, the angle of refraction, θr = 900
The critical angle, θi = ?
We will apply the formula
nisinθi = nrsinθr
Hence, insert the data into the formula
1.52 x sinθi = 1.33 sin900
Divide both sides by 1.52 to obtain
sinθi = (1.33 x sin900) / 1.52 = (1.33 x 1) / 1.52 = 1.33 / 1.52 = 0.875
Hence, sinθi = 0.875
We will make θi subject of the formula
θi = sin-1 0.875 = 61.044980
Therefore, the critical angle θi is 61 degrees.
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