Chapter 7: Q. 111 (page 468)

The index of refraction of light in passing from a vacuum into water is 1.33. If the angle of incidence is 40°, determine the angle of refraction.

Short Answer

Expert verified

The angle of refraction is $28 . 9^{o}$ .

Step by step solution

Step 1. Given Information

The index of refraction of light in passing from a vacuum into water is 1.33. If the angle of incidence is 40°.

We have to determine the angle of refraction.

Step 2. The Snell's law tell us that when you have two medium with refraction index n1 and n2 the equation is 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" role="math" localid="1647100147718" src="https://studysmarter-mediafiles.s3.amazonaws.com/media/textbook-exercise-images/e8ea34e5-f688-4097-b73b-79648fff8fe9.svg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIA4OLDUDE42UZHAIET%2F20220312%2Feu-central-1%2Fs3%2Faws4_request&X-Amz-Date=20220312T155858Z&X-Amz-Expires=90000&X-Amz-SignedHeaders=host&X-Amz-Signature=b1d1d780bcd7cfc5992e5b55f72626883cc0c8fee4ca348b236948ddefa16646" n1sinθ1=n2sinθ2

With $θ_{1} and θ_{2}$ the angle in the first medium and in second medium respectively.

For us the initial angle $θ_{1} = 40^{o} and n_{2} = 1.33$ .

In order to find $θ_{2}$ we need to know $n_{1}$ .

Assume $n_{1} = 1$ for air.

Step 3. Then our equation is

$1 \cdot \sin (40^{o}) = 1.33 \cdot \sin (θ_{2})$

Divide by 1.33 on both side

$\frac{\sin (40^{o})}{1.33} = \frac{1.33}{1.33} \cdot \sin (θ_{2}) \frac{\sin (40^{o})}{1.33} = \sin (θ_{2}) \sin (θ_{2}) = \frac{\sin (40^{o})}{1.33} θ_{2} = \sin^{- 1} (\frac{\sin (40^{o})}{1.33})$

Step 4. After solving the equation is θ2=sin-1sin40o1.33

$θ_{2} \approx \sin^{- 1} (\frac{0.643}{1.33}) θ_{2} \approx \sin^{- 1} (0.483) θ_{2} \approx 28 . 9^{o}$

The angle of refraction is $28 . 9^{o}$ .

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The index of refraction of light in passing from a vacuum into water is 1.33. If the angle of incidence is 40°, determine the angle of refraction.

Short Answer

Step by step solution

Step 1. Given Information

Step 3. Then our equation is

Step 4. After solving the equation is θ2=sin-1sin40o1.33

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