Understanding Refraction - Physics

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Question

A ray of light strikes the surface of a pond at an angle of $30^{\circ}$ to the vertical. If it is moving from air to water, what will be the angle of refraction?

$n_{water}=1.33$

$n_{air}=1.00$

Answer

For this problem, use Snell's law: $n_1\sin(\Theta_1)=n_2\sin(\Theta_2)$ .

In this equation, $\small n$ is the index of refraction and $\Theta$ is the angle of refraction to the vertical. Using the values given in the question, we can find the resultant angle of refraction.

$n_1\sin(\Theta_1)=n_2\sin(\Theta_2)$

$1.00\sin(30^{\circ})=1.33\sin(\Theta_2)$

$0.5=1.33\sin(\Theta_2)$

$\frac{0.5}{1.33}=\sin(\Theta_2)$

$0.376=\sin(\Theta_2)$

Take the arcsin of both sides to find the value of $\Theta_2$ .

$\sin^{-1}(0.376)=\Theta_2$

$22.1^{\circ}=\Theta_2$

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Question

A new crystal is discovered with an index of refraction of . What is the speed of light in this crystal?

$\cmall c=3*10^8 \frac{m}{s}$

Answer

The relationship between speed of light and the index of refraction is:

$n=\frac{c}{v}$

In this formula, is the speed of light in a vacuum, is the observed speed of light in the substance, and is the index of refraction. We are given the speed of light in a vacuum and the index of refraction, allowing us ot solve for the speed of light in the crystal.

$3.33=\frac{310^8\frac{m}{s}}{v}$

$v=\frac{310^8\frac{m}{s}}{3.33}$

$v=9*10^7\frac{m}{s}$

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Question

A new crystal is discovered. The speed of light inside of this crystal is measured to be $210^8\frac{m}{s}$ . What is the index of refraction of the crystal?

$\small c=310^8\frac{m}{s}$

Answer

The relationship between speed of light and the index of refraction is:

$n=\frac{c}{v}$

In this formula, is the speed of light in a vacuum, is the observed speed of light in the substance, and is the index of refraction. We given the values for the speed of light in the crystal and the speed of light in a vacuum, allowing us to solve for the index of refraction.

Plug in our given values and solve.

$n=\frac{310^8\frac{m}{s}}{210^8\frac{m}{s}}$

Note that the index of refraction is a ratio of two velocities, and therefore has no units.

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Question

A certain gas has an index of refraction of . What is the speed of light in the gas?

$\small c=3*10^8\frac{m}{s}$

Answer

The index of refraction is defined as the relationship between the speed of light in a vacuum over the speed of light in the gas/liquid/solid.

$n=\frac{c}{v}$

We know the index of refraction of the gas, and the speed of light in a vacuum. Using these values, we can solve for the speed of light in the gas.

$1.32=\frac{310^8\frac{m}{s}}{v}$

$v=\frac{310^8\frac{m}{s}}{1.32}$

$v=2.27*10^8\frac{m}{s}$

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Question

A beam of light hits a crystal at an angle of $31.5^\circ$ to the normal. The light is refracted in the crystal and changes to an angle of $21.2^\circ$ to the normal. What is the index of refraction of the crystal?

Answer

To solve this problem, use Snell's law:

$n_1\sin(\theta_1)=n_2\sin(\theta_2)$

Remember that air has an index of refraction of . Using this value and the given values for the angle of incidence and angle of refraction, we can find the index of refraction of the crystal.

$n_1\sin(\theta_1)=n_2\sin(\theta_2)$

$1.00*\sin(31.5^\circ)=n_2\sin(21.2^\circ)$

$\frac{\sin(31.5^\circ)}{\sin(21.2^\circ)}=n_2$

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Question

A beam of light hits a crystal at an angle of incidence of $24.3^\circ$ . If the index of refraction of the crystal is , what is the angle of refraction inside of the crystal?

Answer

To solve this problem, use Snell's law:

$n_1\sin(\theta_1)=n_2\sin(\theta_2)$

Remember that air has an index of refraction of . We know the index of refraction for each material and the initial angle of incidence. Using these values, we can find the angle of refraction.

$n_1\sin(\theta_1)=n_2\sin(\theta_2)$

$1.00\sin(24.3^\circ)=1.57\sin(\theta_2)$

$\frac{\sin24.3^\circ}{1.57}=\sin(\theta_2)$

$0.262=\sin(\theta_2)$

$15.2^o=\theta_2$

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Question

Light strikes a crystal from the air at an angle of $12^\circ$ to the normal. Inside the crystal, the angle of light is $8^\circ$ to the normal. What is the index of refraction in the crystal?

Answer

To solve this problem, use Snell's law:

$n_1\sin(\theta_1)=n_2\sin(\theta_2)$

We are given the angle of incidence and the angle of refraction. The index of refraction for air is $\small 1.00$ . Using these values, we can find the index of refraction in the crystal.

$(1.00)\sin(12^\circ)=n_2\sin(8^\circ)$

$\frac{\sin(12^\circ)}{\sin(8^\circ)}=n_2$

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Question

A ray of light strikes the surface of a pond at an angle of $30^{\circ}$ to the vertical. If it is moving from air to water, what will be the angle of refraction?

$n_{water}=1.33$

$n_{air}=1.00$

Answer

For this problem, use Snell's law: $n_1\sin(\Theta_1)=n_2\sin(\Theta_2)$ .

In this equation, $\small n$ is the index of refraction and $\Theta$ is the angle of refraction to the vertical. Using the values given in the question, we can find the resultant angle of refraction.

$n_1\sin(\Theta_1)=n_2\sin(\Theta_2)$

$1.00\sin(30^{\circ})=1.33\sin(\Theta_2)$

$0.5=1.33\sin(\Theta_2)$

$\frac{0.5}{1.33}=\sin(\Theta_2)$

$0.376=\sin(\Theta_2)$

Take the arcsin of both sides to find the value of $\Theta_2$ .

$\sin^{-1}(0.376)=\Theta_2$

$22.1^{\circ}=\Theta_2$

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Question

A new crystal is discovered with an index of refraction of . What is the speed of light in this crystal?

$\cmall c=3*10^8 \frac{m}{s}$

Answer

The relationship between speed of light and the index of refraction is:

$n=\frac{c}{v}$

In this formula, is the speed of light in a vacuum, is the observed speed of light in the substance, and is the index of refraction. We are given the speed of light in a vacuum and the index of refraction, allowing us ot solve for the speed of light in the crystal.

$3.33=\frac{310^8\frac{m}{s}}{v}$

$v=\frac{310^8\frac{m}{s}}{3.33}$

$v=9*10^7\frac{m}{s}$

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Question

A new crystal is discovered. The speed of light inside of this crystal is measured to be $210^8\frac{m}{s}$ . What is the index of refraction of the crystal?

$\small c=310^8\frac{m}{s}$

Answer

The relationship between speed of light and the index of refraction is:

$n=\frac{c}{v}$

In this formula, is the speed of light in a vacuum, is the observed speed of light in the substance, and is the index of refraction. We given the values for the speed of light in the crystal and the speed of light in a vacuum, allowing us to solve for the index of refraction.

Plug in our given values and solve.

$n=\frac{310^8\frac{m}{s}}{210^8\frac{m}{s}}$

Note that the index of refraction is a ratio of two velocities, and therefore has no units.

Compare your answer with the correct one above

Question

A certain gas has an index of refraction of . What is the speed of light in the gas?

$\small c=3*10^8\frac{m}{s}$

Answer

The index of refraction is defined as the relationship between the speed of light in a vacuum over the speed of light in the gas/liquid/solid.

$n=\frac{c}{v}$

We know the index of refraction of the gas, and the speed of light in a vacuum. Using these values, we can solve for the speed of light in the gas.

$1.32=\frac{310^8\frac{m}{s}}{v}$

$v=\frac{310^8\frac{m}{s}}{1.32}$

$v=2.27*10^8\frac{m}{s}$

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Question

A beam of light hits a crystal at an angle of $31.5^\circ$ to the normal. The light is refracted in the crystal and changes to an angle of $21.2^\circ$ to the normal. What is the index of refraction of the crystal?

Answer

To solve this problem, use Snell's law:

$n_1\sin(\theta_1)=n_2\sin(\theta_2)$

Remember that air has an index of refraction of . Using this value and the given values for the angle of incidence and angle of refraction, we can find the index of refraction of the crystal.

$n_1\sin(\theta_1)=n_2\sin(\theta_2)$

$1.00*\sin(31.5^\circ)=n_2\sin(21.2^\circ)$

$\frac{\sin(31.5^\circ)}{\sin(21.2^\circ)}=n_2$

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Question

A beam of light hits a crystal at an angle of incidence of $24.3^\circ$ . If the index of refraction of the crystal is , what is the angle of refraction inside of the crystal?

Answer

To solve this problem, use Snell's law:

$n_1\sin(\theta_1)=n_2\sin(\theta_2)$

Remember that air has an index of refraction of . We know the index of refraction for each material and the initial angle of incidence. Using these values, we can find the angle of refraction.

$n_1\sin(\theta_1)=n_2\sin(\theta_2)$

$1.00\sin(24.3^\circ)=1.57\sin(\theta_2)$

$\frac{\sin24.3^\circ}{1.57}=\sin(\theta_2)$

$0.262=\sin(\theta_2)$

$15.2^o=\theta_2$

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Question

Light strikes a crystal from the air at an angle of $12^\circ$ to the normal. Inside the crystal, the angle of light is $8^\circ$ to the normal. What is the index of refraction in the crystal?

Answer

To solve this problem, use Snell's law:

$n_1\sin(\theta_1)=n_2\sin(\theta_2)$

We are given the angle of incidence and the angle of refraction. The index of refraction for air is $\small 1.00$ . Using these values, we can find the index of refraction in the crystal.

$(1.00)\sin(12^\circ)=n_2\sin(8^\circ)$

$\frac{\sin(12^\circ)}{\sin(8^\circ)}=n_2$

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Question

A ray of light strikes the surface of a pond at an angle of $30^{\circ}$ to the vertical. If it is moving from air to water, what will be the angle of refraction?

$n_{water}=1.33$

$n_{air}=1.00$

Answer

For this problem, use Snell's law: $n_1\sin(\Theta_1)=n_2\sin(\Theta_2)$ .

In this equation, $\small n$ is the index of refraction and $\Theta$ is the angle of refraction to the vertical. Using the values given in the question, we can find the resultant angle of refraction.

$n_1\sin(\Theta_1)=n_2\sin(\Theta_2)$

$1.00\sin(30^{\circ})=1.33\sin(\Theta_2)$

$0.5=1.33\sin(\Theta_2)$

$\frac{0.5}{1.33}=\sin(\Theta_2)$

$0.376=\sin(\Theta_2)$

Take the arcsin of both sides to find the value of $\Theta_2$ .

$\sin^{-1}(0.376)=\Theta_2$

$22.1^{\circ}=\Theta_2$

Compare your answer with the correct one above

Question

A new crystal is discovered with an index of refraction of . What is the speed of light in this crystal?

$\cmall c=3*10^8 \frac{m}{s}$

Answer

The relationship between speed of light and the index of refraction is:

$n=\frac{c}{v}$

In this formula, is the speed of light in a vacuum, is the observed speed of light in the substance, and is the index of refraction. We are given the speed of light in a vacuum and the index of refraction, allowing us ot solve for the speed of light in the crystal.

$3.33=\frac{310^8\frac{m}{s}}{v}$

$v=\frac{310^8\frac{m}{s}}{3.33}$

$v=9*10^7\frac{m}{s}$

Compare your answer with the correct one above

Question

A new crystal is discovered. The speed of light inside of this crystal is measured to be $210^8\frac{m}{s}$ . What is the index of refraction of the crystal?

$\small c=310^8\frac{m}{s}$

Answer

The relationship between speed of light and the index of refraction is:

$n=\frac{c}{v}$

In this formula, is the speed of light in a vacuum, is the observed speed of light in the substance, and is the index of refraction. We given the values for the speed of light in the crystal and the speed of light in a vacuum, allowing us to solve for the index of refraction.

Plug in our given values and solve.

$n=\frac{310^8\frac{m}{s}}{210^8\frac{m}{s}}$

Note that the index of refraction is a ratio of two velocities, and therefore has no units.

Compare your answer with the correct one above

Question

A certain gas has an index of refraction of . What is the speed of light in the gas?

$\small c=3*10^8\frac{m}{s}$

Answer

The index of refraction is defined as the relationship between the speed of light in a vacuum over the speed of light in the gas/liquid/solid.

$n=\frac{c}{v}$

We know the index of refraction of the gas, and the speed of light in a vacuum. Using these values, we can solve for the speed of light in the gas.

$1.32=\frac{310^8\frac{m}{s}}{v}$

$v=\frac{310^8\frac{m}{s}}{1.32}$

$v=2.27*10^8\frac{m}{s}$

Compare your answer with the correct one above

Question

A beam of light hits a crystal at an angle of $31.5^\circ$ to the normal. The light is refracted in the crystal and changes to an angle of $21.2^\circ$ to the normal. What is the index of refraction of the crystal?

Answer

To solve this problem, use Snell's law:

$n_1\sin(\theta_1)=n_2\sin(\theta_2)$

Remember that air has an index of refraction of . Using this value and the given values for the angle of incidence and angle of refraction, we can find the index of refraction of the crystal.

$n_1\sin(\theta_1)=n_2\sin(\theta_2)$

$1.00*\sin(31.5^\circ)=n_2\sin(21.2^\circ)$

$\frac{\sin(31.5^\circ)}{\sin(21.2^\circ)}=n_2$

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Question

A beam of light hits a crystal at an angle of incidence of $24.3^\circ$ . If the index of refraction of the crystal is , what is the angle of refraction inside of the crystal?

Answer

To solve this problem, use Snell's law:

$n_1\sin(\theta_1)=n_2\sin(\theta_2)$

Remember that air has an index of refraction of . We know the index of refraction for each material and the initial angle of incidence. Using these values, we can find the angle of refraction.

$n_1\sin(\theta_1)=n_2\sin(\theta_2)$

$1.00\sin(24.3^\circ)=1.57\sin(\theta_2)$

$\frac{\sin24.3^\circ}{1.57}=\sin(\theta_2)$

$0.262=\sin(\theta_2)$

$15.2^o=\theta_2$

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