refraction worksheet
1 calculate the speed of light in (a) fluorite (b) cubic zirconia (c) diamond.

2. the speed of light in a certain plastic is 2.00 x 10⁸ m/s. what is the index of refraction of this plastic?

3. light is incident upon a piece of crown glass from air at a 45° angle. (a) what is the angle of refraction? (b) some of the light is reflected from the glass. what is the angle of reflection?

4. a light ray travels from air into a liquid. the ray is incident to the liquid at 30° and the angle of refraction is 22°. what is the index of refraction of the liquid?

5. a ray of light in flint glass is incident onto water at an angle of 24°. at what angle does the light ray enter the water?

6. light travels through water at 2.2506 x 10⁸ m/s. a person shines a light at a friend that is underwater. if the ray in the water makes an angle of 38° with the normal, what is the angle of incidence?

7. use the following diagram and calculate the θ₁ and θ₂. the index of refraction for oil is 1.48.
(diagram: air, oil, water layers with light ray, angles. θ₁ at air-oil interface, 20° between oil normal and ray, θ₂ at oil-water interface)

fluorite n = 1.434
crown glass n = 1.52
cubic zirconia n = 2.20
diamond n = 2.419
flint glass n = 1.66
water n = 1.333

Question

refraction worksheet
1 calculate the speed of light in (a) fluorite (b) cubic zirconia (c) diamond.

2. the speed of light in a certain plastic is 2.00 x 10⁸ m/s. what is the index of refraction of this plastic?

3. light is incident upon a piece of crown glass from air at a 45° angle. (a) what is the angle of refraction? (b) some of the light is reflected from the glass. what is the angle of reflection?

4. a light ray travels from air into a liquid. the ray is incident to the liquid at 30° and the angle of refraction is 22°. what is the index of refraction of the liquid?

5. a ray of light in flint glass is incident onto water at an angle of 24°. at what angle does the light ray enter the water?

6. light travels through water at 2.2506 x 10⁸ m/s. a person shines a light at a friend that is underwater. if the ray in the water makes an angle of 38° with the normal, what is the angle of incidence?

7. use the following diagram and calculate the θ₁ and θ₂. the index of refraction for oil is 1.48.
(diagram: air, oil, water layers with light ray, angles. θ₁ at air-oil interface, 20° between oil normal and ray, θ₂ at oil-water interface)

fluorite n = 1.434
crown glass n = 1.52
cubic zirconia n = 2.20
diamond n = 2.419
flint glass n = 1.66
water n = 1.333

Accepted Answer

Let's solve problem 2 first as an example.

# Explanation:
## Step1: Recall the formula for refractive index
The formula for the refractive index $ n $ of a medium is $ n=\frac{c}{v} $, where $ c $ is the speed of light in vacuum ($ c = 3.00\times10^{8}\space m/s $) and $ v $ is the speed of light in the medium.
## Step2: Substitute the given values
We are given that $ v = 2.00\times10^{8}\space m/s $ and $ c=3.00\times10^{8}\space m/s $. Substituting these values into the formula:
$ n=\frac{3.00\times 10^{8}}{2.00\times 10^{8}} $
## Step3: Calculate the value
Simplifying the above expression, the $ 10^{8} $ terms cancel out, and we have $ n = \frac{3.00}{2.00}=1.5 $

# Answer:
The index of refraction of the plastic is $ 1.5 $

Now, let's solve problem 1 (part A: fluorite)
# Explanation:
## Step1: Recall the formula for speed of light in a medium
The formula is $ v=\frac{c}{n} $, where $ c = 3.00\times10^{8}\space m/s $ and $ n $ is the refractive index of the medium. From the table, the refractive index of fluorite $ n = 1.434 $
## Step2: Substitute the values into the formula
$ v=\frac{3.00\times 10^{8}}{1.434} $
## Step3: Calculate the value
$ v\approx2.09\times 10^{8}\space m/s $ (rounded to three significant figures)

# Answer:
The speed of light in fluorite is approximately $ 2.09\times 10^{8}\space m/s $

For part B of problem 1 (cubic zirconia):
# Explanation:
## Step1: Use the same formula $ v=\frac{c}{n} $
From the table, the refractive index of cubic zirconia $ n = 2.20 $ and $ c = 3.00\times10^{8}\space m/s $
## Step2: Substitute the values
$ v=\frac{3.00\times 10^{8}}{2.20} $
## Step3: Calculate the value
$ v\approx1.36\times 10^{8}\space m/s $ (rounded to three significant figures)

# Answer:
The speed of light in cubic zirconia is approximately $ 1.36\times 10^{8}\space m/s $

For part C of problem 1 (diamond):
# Explanation:
## Step1: Use the formula $ v=\frac{c}{n} $
From the table, the refractive index of diamond $ n = 2.419 $ and $ c = 3.00\times10^{8}\space m/s $
## Step2: Substitute the values
$ v=\frac{3.00\times 10^{8}}{2.419} $
## Step3: Calculate the value
$ v\approx1.24\times 10^{8}\space m/s $ (rounded to three significant figures)

# Answer:
The speed of light in diamond is approximately $ 1.24\times 10^{8}\space m/s $

For problem 3 (A: angle of refraction in crown glass):
# Explanation:
## Step1: Recall Snell's Law
Snell's Law is $ n_1\sin\theta_1=n_2\sin\theta_2 $, where $ n_1 $ is the refractive index of the incident medium (air, $ n_1 = 1 $), $ \theta_1 $ is the angle of incidence ($ 45^\circ $), $ n_2 $ is the refractive index of crown glass ($ n_2=1.52 $), and $ \theta_2 $ is the angle of refraction.
## Step2: Rearrange Snell's Law to solve for $ \theta_2 $
$ \sin\theta_2=\frac{n_1\sin\theta_1}{n_2} $
Substitute $ n_1 = 1 $, $ \theta_1 = 45^\circ $, $ n_2 = 1.52 $
$ \sin\theta_2=\frac{1\times\sin(45^\circ)}{1.52} $
$ \sin(45^\circ)=\frac{\sqrt{2}}{2}\approx0.7071 $
So, $ \sin\theta_2=\frac{0.7071}{1.52}\approx0.4652 $
## Step3: Find $ \theta_2 $
$ \theta_2=\arcsin(0.4652)\approx27.7^\circ $ (rounded to three significant figures)

# Answer:
The angle of refraction is approximately $ 27.7^\circ $

For problem 3 (B: angle of reflection):
# Explanation:
## Step1: Recall the Law of Reflection
The Law of Reflection states that the angle of incidence is equal to the angle of reflection, i.e., $ \theta_{reflection}=\theta_{incidence} $
## Step2: Substitute the given angle of incidence
The angle of incidence is $ 45^\circ $, so the angle of reflection is also $ 45^\circ $

# Answer:
The angle of reflection is $ 45^\circ $

For problem 4:
# Explanation:
## Step1: Apply Snell's Law
Snell's Law: $ n_1\sin\theta_1=n_2\sin\theta_2 $, where $ n_1 = 1 $ (air), $ \theta_1 = 30^\circ $, $ \theta_2 = 22^\circ $, and $ n_2 $ is the refractive index of the liquid.
## Step2: Solve for $ n_2 $
$ n_2=\frac{n_1\sin\theta_1}{\sin\theta_2} $
Substitute $ n_1 = 1 $, $ \theta_1 = 30^\circ $, $ \theta_2 = 22^\circ $
$ \sin(30^\circ)=0.5 $, $ \sin(22^\circ)\approx0.3746 $
$ n_2=\frac{1\times0.5}{0.3746}\approx1.33 $

# Answer:
The index of refraction of the liquid is approximately $ 1.33 $

For problem 5:
# Explanation:
## Step1: Identify the refractive indices
Refractive index of flint glass $ n_1 = 1.66 $, refractive index of water $ n_2 = 1.333 $, angle of incidence in flint glass $ \theta_1 = 24^\circ $
## Step2: Apply Snell's Law $ n_1\sin\theta_1=n_2\sin\theta_2 $
Solve for $ \theta_2 $ (angle of refraction in water, which is the angle at which light enters water)
$ \sin\theta_2=\frac{n_1\sin\theta_1}{n_2} $
$ \sin(24^\circ)\approx0.4067 $
$ \sin\theta_2=\frac{1.66\times0.4067}{1.333}\approx\frac{0.6751}{1.333}\approx0.5065 $
## Step3: Find $ \theta_2 $
$ \theta_2=\arcsin(0.5065)\approx30.4^\circ $ (rounded to three significant figures)

# Answer:
The light ray enters water at an angle of approximately $ 30.4^\circ $

For problem 6:
# Explanation:
## Step1: First, find the refractive index of water
We know that $ v=\frac{c}{n} $, so $ n=\frac{c}{v} $. Given $ v = 2.2506\times10^{8}\space m/s $ and $ c = 3.00\times10^{8}\space m/s $
$ n=\frac{3.00\times 10^{8}}{2.2506\times 10^{8}}\approx1.333 $ (which matches the given value for water's refractive index)
## Step2: Apply Snell's Law
Snell's Law: $ n_1\sin\theta_1=n_2\sin\theta_2 $, where $ n_1 = 1 $ (air), $ \theta_1 $ is the angle of incidence (what we need to find), $ n_2 = 1.333 $ (water), and $ \theta_2 = 38^\circ $
## Step3: Solve for $ \theta_1 $
$ \sin\theta_1=\frac{n_2\sin\theta_2}{n_1} $
Substitute $ n_1 = 1 $, $ n_2 = 1.333 $, $ \theta_2 = 38^\circ $
$ \sin(38^\circ)\approx0.6157 $
$ \sin\theta_1=\frac{1.333\times0.6157}{1}\approx0.8207 $
## Step4: Find $ \theta_1 $
$ \theta_1=\arcsin(0.8207)\approx55.1^\circ $ (rounded to three significant figures)

# Answer:
The angle of incidence is approximately $ 55.1^\circ $

For problem 7:
First, let's assume the diagram has two interfaces: air - oil and oil - water. Let's assume the angle between the normal and the oil - water interface ray is related to $ \theta_2 $, but from the diagram, we see that the angle between the normal and the oil ray is $ 20^\circ $, so that's the angle of refraction when going from air to oil? Wait, no, let's re - examine.

Wait, the first interface is air - oil. Let's assume the ray is going from air to oil first. Wait, the diagram shows a ray in air with angle $ \theta_1 $ with the normal, then refracts into oil with angle $ 20^\circ $ with the normal, then refracts into water with angle $ \theta_2 $ with the normal.

First, for air - oil interface:

# Explanation (for $ \theta_1 $):
## Step1: Apply Snell's Law for air - oil
$ n_{air}\sin\theta_1=n_{oil}\sin\theta_{oil} $
$ n_{air}=1 $, $ n_{oil}=1.48 $, $ \theta_{oil}=20^\circ $
## Step2: Solve for $ \theta_1 $
$ \sin\theta_1=\frac{n_{oil}\sin\theta_{oil}}{n_{air}} $
$ \sin(20^\circ)\approx0.3420 $
$ \sin\theta_1 = 1.48\times0.3420\approx0.5062 $
## Step3: Find $ \theta_1 $
$ \theta_1=\arcsin(0.5062)\approx30.4^\circ $

Now, for oil - water interface:

## Step1: Apply Snell's Law for oil - water
$ n_{oil}\sin\theta_{oil}=n_{water}\sin\theta_2 $
$ n_{oil}=1.48 $, $ \theta_{oil}=20^\circ $, $ n_{water}=1.333 $
## Step2: Solve for $ \theta_2 $
$ \sin\theta_2=\frac{n_{oil}\sin\theta_{oil}}{n_{water}} $
$ \sin\theta_2=\frac{1.48\times0.3420}{1.333}\approx\frac{0.5062}{1.333}\approx0.3797 $
## Step3: Find $ \theta_2 $
$ \theta_2=\arcsin(0.3797)\approx22.3^\circ $

# Answer (for $ \theta_1 $):
$ \theta_1\approx30.4^\circ $

# Answer (for $ \theta_2 $):
$ \theta_2\approx22.3^\circ $

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the formula for refractive index

Step2: Substitute the given values

Step3: Calculate the value

Step1: Recall the formula for speed of light in a medium

Step2: Substitute the values into the formula

Step3: Calculate the value

Step1: Use the same formula \( v=\frac{c}{n} \)

Step2: Substitute the values

Step3: Calculate the value

Answer: