## Question

What is the critical angle for light traveling from crown glass (n = 1.52) into water (n = 1.33)?

## Answer

**The final answer to the above question is that the critical angle is 61 degrees (61 ^{0}) **

### Proof

Here is the method I applied to solve the problem:

**Data**

We will identify our data from the above question

The light travels from crown glass = Incident Light = n_{i} = 1.52

It then passes through water = Refracted ray of light = n_{r} = 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 90^{0}. Therefore, the angle of refraction, θ_{r} = 90^{0}

**Unknown**

The critical angle, θ_{i} = ?

**Formula**

We will apply the formula

n_{i}sinθ_{i} = n_{r}sinθ_{r}

#### Solution

Hence, insert the data into the formula

1.52 x sinθ_{i} = 1.33 sin90^{0}

Divide both sides by 1.52 to obtain

sinθ_{i} = (1.33 x sin90^{0}) / 1.52 = (1.33 x 1) / 1.52 = 1.33 / 1.52 = 0.875

Hence, sinθ_{i} = 0.875

We will make θ_{i} subject of the formula

θ_{i} = sin^{-1} 0.875 = 61.04498^{0}

**Therefore, the critical angle θ _{i} is 61 degrees.**

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