Other Optics Principles - AP Physics 2

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Question

Light rays encounter a mystery optical device, resulting in a new distribution of the light waves as shown. Assume the light travels from right to left.

What type of reflecting or refracting surface is depicted?

Answer

In a converging lens, the light waves pass through it and have their angles altered so that they point closer together than they did before they went through the lens. In the picture, the light waves are diverging from a point until they go into the lens, at which point they no longer diverge from each other. Therefore, this is a converging lens. Because the waves are travelling the same direction the whole time, it can't be the converging or diverging mirrors. If the lens were diverging, they'd be more separated. If it were a plane mirror, the waves would get polarized (they'd have the same phase angle).

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Question

If a person with near point distance of 20cm observes a fine detailed coin with magnifying glass with an angular magnification of 5, what is the focal length?

Answer

Use the formula for focal length.

$M=\frac{\frac{h_0}{f}}{\frac{h_0}{N}}$

Solve for focal length by substituting known values.

$f=\frac{N}{M} = \frac{20 { {cm}}}{5}=4 { cm}=0.04{ m}$

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Question

Malus' law:

$I=I_0cos^2(\theta)$

Where is the intensity of polarized light that has passed through the polarizer, is the intensity of polarized light before the polarizer, and $\theta$ is the angle between the polarized light and the polarizer.

Unpolarized light passes through a polarizer. It then passes through another at angle $25^{\circ}$ to the first. What percentage of the original intensity was the light coming out of the second polarizer?

Answer

Using Malus' law.

$I=I_0cos^2(\theta)$

Since the initial light is unpolarized, there will be no intensity lost.

$I=I_0cos^2(25^{\circ})$

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Question

Malus' law:

$I=I_0cos^2(\theta)$

Where is the intensity of polarized light that has passed through the polarizer, is the intensity of polarized light before the polarizer, and $\theta$ is the angle between the polarized light and the polarizer.

Unpolarized light passes through a polarizer. It then passes through another polarizer at angle $35^{\circ}$ to the first, and then another at angle $25^{\circ}$ to the second. What percentage of the original intensity was the light coming out of the third polarizer?

Answer

Use Malus' law.

$I=I_0cos^2(\theta)$

The light's intensity is reduced by the final two polarizers. It is thus necessary to use Malus' law twice.

$I_1=I_0cos^2(\theta_1)$

$I_2=I_1cos^2(\theta_2)$

Where is the initial intensity after the first polarizer.

is the intensity after the second polarizer.

is the intensity after the third polarizer.

$\theta_1$ is the angle between the first and second polarizers.

$\theta_2$ is the angle between the second and third polarizers.

Combining equations:

$I_2=I_0cos^2(\theta_2)cos^2({\theta_1})$

Plug in values:

$I_2=I_0cos^2(25^\circ)cos^2(35^\circ)$

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Question

Malus' law:

$I=I_0cos^2(\theta)$

Where is the intensity of polarized light that has passed through the polarizer, is the intensity of polarized light before the polarizer, and $\theta$ is the angle between the polarized light and the polarizer.

Unpolarized light passes through a polarizer. It then passes through another polarizer at angle $55^{\circ}$ to the first, and then another at angle $65^{\circ}$ to the second. What percentage of the original intensity was the light coming out of the third polarizer?

Answer

Use Malus' law.

$I=I_0cos^2(\theta)$

The light's intensity is reduced by the final two polarizers. We will need to use Malus' law twice.

$I_1=I_0cos^2(\theta_1)$

$I_2=I_1cos^2(\theta_2)$

Where is the initial intensity after the first polarizer.

is the intensity after the second polarizer.

is the intensity after the third polarizer.

$\theta_1$ is the angle between the first and second polarizers.

$\theta_2$ is the angle between the second and third polarizers.

Combining equations:

$I_2=I_0cos^2(\theta_2)cos^2({\theta_1})$

Plug in values:

$I_2=I_0cos^2(65^\circ)cos^2(55^\circ)$

Compare your answer with the correct one above

Question

Malus' law:

$I=I_0cos^2(\theta)$

Where is the intensity of polarized light that has passed through the polarizer, is the intensity of polarized light before the polarizer, and $\theta$ is the angle between the polarized light and the polarizer.

Unpolarized light passes through a polarizer. It then passes through another polarizer at angle $5^{\circ}$ to the first, and then another at angle $45^{\circ}$ to the second. What percentage of the original intensity was the light coming out of the third polarizer?

Answer

Use Malus' law.

$I=I_0cos^2(\theta)$

The light's intensity is reduced by the final two polarizers. We will need to use Malus' law twice.

$I_1=I_0cos^2(\theta_1)$

$I_2=I_1cos^2(\theta_2)$

Where is the initial intensity after the first polarizer.

is the intensity after the second polarizer.

is the intensity after the third polarizer.

$\theta_1$ is the angle between the first and second polarizers.

$\theta_2$ is the angle between the second and third polarizers.

Combine equations:

$I_2=I_0cos^2(\theta_2)cos^2({\theta_1})$

Plug in values:

$I_2=I_0cos^2(45^\circ)cos^2(5^\circ)$

Compare your answer with the correct one above

Question

Malus' law:

$I=I_0cos^2(\theta)$

Where is the intensity of polarized light that has passed through the polarizer, is the intensity of polarized light before the polarizer, and $\theta$ is the angle between the polarized light and the polarizer.

Unpolarized light passes through a polarizer. It then passes through another polarizer at angle $40^{\circ}$ to the first, and then another at angle $15^{\circ}$ to the second. What percentage of the original intensity was the light coming out of the second polarizer?

Answer

Use Malus' law.

$I=I_0cos^2(\theta)$

The light's intensity is reduced by the final two polarizers. We will need to use Malus' law twice.

$I_1=I_0cos^2(\theta_1)$

$I_2=I_1cos^2(\theta_2)$

Where is the initial intensity after the first polarizer.

is the intensity after the second polarizer.

is the intensity after the third polarizer.

$\theta_1$ is the angle between the first and second polarizers.

$\theta_2$ is the angle between the second and third polarizers.

Combine equations:

$I_2=I_0cos^2(\theta_2)cos^2({\theta_1})$

Plug in values:

$I_2=I_0cos^2(15^\circ)cos^2({40^\circ})$

Compare your answer with the correct one above

Question

Malus' law:

$I=I_0cos^2(\theta)$

Where is the intensity of polarized light that has passed through the polarizer, is the intensity of polarized light before the polarizer, and $\theta$ is the angle between the polarized light and the polarizer.

Unpolarized light passes through a polarizer. It then passes through another polarizer at angle $30^{\circ}$ to the first, and then another at angle $50^{\circ}$ to the second. What percentage of the original intensity was the light coming out of the third polarizer?

Answer

Use Malus' law.

$I=I_0cos^2(\theta)$

The light's intensity is reduced by the final two polarizers. We will need to use Malus law twice.

$I_1=I_0cos^2(\theta_1)$

$I_2=I_1cos^2(\theta_2)$

Where is the initial intensity after the first polarizer.

is the intensity after the second polarizer.

is the intensity after the third polarizer.

$\theta_1$ is the angle between the first and second polarizers.

$\theta_2$ is the angle between the second and third polarizers.

Combine equations:

$I_2=I_0cos^2(\theta_2)cos^2({\theta_1})$

Plug in values:

$I_2=I_0cos^2(30^\circ)cos^2(50^\circ)$

Compare your answer with the correct one above

Question

You are passing a ray of light through clear alcohol to determine properties. You shine the light ray exactly to the surface of alcohol.

Suppose that the alcohol surface will completely reflect the incoming light beam like a mirror. At what angle will the angle of reflectance be to the normal of the surface of the alcohol?

Answer

Assuming the alcohol works like a perfect mirror, the angle of reflectance will equal the angle of incidence. Therefore, the angle of reflectance will be

Compare your answer with the correct one above

Question

The diffraction of waves refers to the

Answer

Diffraction is the bending of light around an object. It is seen in the Double Slit experiment in which waves are spread out from their initial line of travel.

Compare your answer with the correct one above

Question

Light rays encounter a mystery optical device, resulting in a new distribution of the light waves as shown. Assume the light travels from right to left.

What type of reflecting or refracting surface is depicted?

Answer

In a converging lens, the light waves pass through it and have their angles altered so that they point closer together than they did before they went through the lens. In the picture, the light waves are diverging from a point until they go into the lens, at which point they no longer diverge from each other. Therefore, this is a converging lens. Because the waves are travelling the same direction the whole time, it can't be the converging or diverging mirrors. If the lens were diverging, they'd be more separated. If it were a plane mirror, the waves would get polarized (they'd have the same phase angle).

Compare your answer with the correct one above

Question

If a person with near point distance of 20cm observes a fine detailed coin with magnifying glass with an angular magnification of 5, what is the focal length?

Answer

Use the formula for focal length.

$M=\frac{\frac{h_0}{f}}{\frac{h_0}{N}}$

Solve for focal length by substituting known values.

$f=\frac{N}{M} = \frac{20 { {cm}}}{5}=4 { cm}=0.04{ m}$

Compare your answer with the correct one above

Question

Malus' law:

$I=I_0cos^2(\theta)$

Where is the intensity of polarized light that has passed through the polarizer, is the intensity of polarized light before the polarizer, and $\theta$ is the angle between the polarized light and the polarizer.

Unpolarized light passes through a polarizer. It then passes through another at angle $25^{\circ}$ to the first. What percentage of the original intensity was the light coming out of the second polarizer?

Answer

Using Malus' law.

$I=I_0cos^2(\theta)$

Since the initial light is unpolarized, there will be no intensity lost.

$I=I_0cos^2(25^{\circ})$

Compare your answer with the correct one above

Question

Malus' law:

$I=I_0cos^2(\theta)$

Where is the intensity of polarized light that has passed through the polarizer, is the intensity of polarized light before the polarizer, and $\theta$ is the angle between the polarized light and the polarizer.

Unpolarized light passes through a polarizer. It then passes through another polarizer at angle $35^{\circ}$ to the first, and then another at angle $25^{\circ}$ to the second. What percentage of the original intensity was the light coming out of the third polarizer?

Answer

Use Malus' law.

$I=I_0cos^2(\theta)$

The light's intensity is reduced by the final two polarizers. It is thus necessary to use Malus' law twice.

$I_1=I_0cos^2(\theta_1)$

$I_2=I_1cos^2(\theta_2)$

Where is the initial intensity after the first polarizer.

is the intensity after the second polarizer.

is the intensity after the third polarizer.

$\theta_1$ is the angle between the first and second polarizers.

$\theta_2$ is the angle between the second and third polarizers.

Combining equations:

$I_2=I_0cos^2(\theta_2)cos^2({\theta_1})$

Plug in values:

$I_2=I_0cos^2(25^\circ)cos^2(35^\circ)$

Compare your answer with the correct one above

Question

Malus' law:

$I=I_0cos^2(\theta)$

Where is the intensity of polarized light that has passed through the polarizer, is the intensity of polarized light before the polarizer, and $\theta$ is the angle between the polarized light and the polarizer.

Unpolarized light passes through a polarizer. It then passes through another polarizer at angle $55^{\circ}$ to the first, and then another at angle $65^{\circ}$ to the second. What percentage of the original intensity was the light coming out of the third polarizer?

Answer

Use Malus' law.

$I=I_0cos^2(\theta)$

The light's intensity is reduced by the final two polarizers. We will need to use Malus' law twice.

$I_1=I_0cos^2(\theta_1)$

$I_2=I_1cos^2(\theta_2)$

Where is the initial intensity after the first polarizer.

is the intensity after the second polarizer.

is the intensity after the third polarizer.

$\theta_1$ is the angle between the first and second polarizers.

$\theta_2$ is the angle between the second and third polarizers.

Combining equations:

$I_2=I_0cos^2(\theta_2)cos^2({\theta_1})$

Plug in values:

$I_2=I_0cos^2(65^\circ)cos^2(55^\circ)$

Compare your answer with the correct one above

Question

Malus' law:

$I=I_0cos^2(\theta)$

Where is the intensity of polarized light that has passed through the polarizer, is the intensity of polarized light before the polarizer, and $\theta$ is the angle between the polarized light and the polarizer.

Unpolarized light passes through a polarizer. It then passes through another polarizer at angle $5^{\circ}$ to the first, and then another at angle $45^{\circ}$ to the second. What percentage of the original intensity was the light coming out of the third polarizer?

Answer

Use Malus' law.

$I=I_0cos^2(\theta)$

The light's intensity is reduced by the final two polarizers. We will need to use Malus' law twice.

$I_1=I_0cos^2(\theta_1)$

$I_2=I_1cos^2(\theta_2)$

Where is the initial intensity after the first polarizer.

is the intensity after the second polarizer.

is the intensity after the third polarizer.

$\theta_1$ is the angle between the first and second polarizers.

$\theta_2$ is the angle between the second and third polarizers.

Combine equations:

$I_2=I_0cos^2(\theta_2)cos^2({\theta_1})$

Plug in values:

$I_2=I_0cos^2(45^\circ)cos^2(5^\circ)$

Compare your answer with the correct one above

Question

Malus' law:

$I=I_0cos^2(\theta)$

Where is the intensity of polarized light that has passed through the polarizer, is the intensity of polarized light before the polarizer, and $\theta$ is the angle between the polarized light and the polarizer.

Unpolarized light passes through a polarizer. It then passes through another polarizer at angle $40^{\circ}$ to the first, and then another at angle $15^{\circ}$ to the second. What percentage of the original intensity was the light coming out of the second polarizer?

Answer

Use Malus' law.

$I=I_0cos^2(\theta)$

The light's intensity is reduced by the final two polarizers. We will need to use Malus' law twice.

$I_1=I_0cos^2(\theta_1)$

$I_2=I_1cos^2(\theta_2)$

Where is the initial intensity after the first polarizer.

is the intensity after the second polarizer.

is the intensity after the third polarizer.

$\theta_1$ is the angle between the first and second polarizers.

$\theta_2$ is the angle between the second and third polarizers.

Combine equations:

$I_2=I_0cos^2(\theta_2)cos^2({\theta_1})$

Plug in values:

$I_2=I_0cos^2(15^\circ)cos^2({40^\circ})$

Compare your answer with the correct one above

Question

Malus' law:

$I=I_0cos^2(\theta)$

Where is the intensity of polarized light that has passed through the polarizer, is the intensity of polarized light before the polarizer, and $\theta$ is the angle between the polarized light and the polarizer.

Unpolarized light passes through a polarizer. It then passes through another polarizer at angle $30^{\circ}$ to the first, and then another at angle $50^{\circ}$ to the second. What percentage of the original intensity was the light coming out of the third polarizer?

Answer

Use Malus' law.

$I=I_0cos^2(\theta)$

The light's intensity is reduced by the final two polarizers. We will need to use Malus law twice.

$I_1=I_0cos^2(\theta_1)$

$I_2=I_1cos^2(\theta_2)$

Where is the initial intensity after the first polarizer.

is the intensity after the second polarizer.

is the intensity after the third polarizer.

$\theta_1$ is the angle between the first and second polarizers.

$\theta_2$ is the angle between the second and third polarizers.

Combine equations:

$I_2=I_0cos^2(\theta_2)cos^2({\theta_1})$

Plug in values:

$I_2=I_0cos^2(30^\circ)cos^2(50^\circ)$

Compare your answer with the correct one above

Question

You are passing a ray of light through clear alcohol to determine properties. You shine the light ray exactly to the surface of alcohol.

Suppose that the alcohol surface will completely reflect the incoming light beam like a mirror. At what angle will the angle of reflectance be to the normal of the surface of the alcohol?

Answer

Assuming the alcohol works like a perfect mirror, the angle of reflectance will equal the angle of incidence. Therefore, the angle of reflectance will be

Compare your answer with the correct one above

Question

The diffraction of waves refers to the

Answer

Diffraction is the bending of light around an object. It is seen in the Double Slit experiment in which waves are spread out from their initial line of travel.

Compare your answer with the correct one above

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