General Principles and Properties - MCAT Physical

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Question

A light ray traveling through a medium is reflected by a second medium at an angle of 20⁰ to the interface between the two media. Which of the following is true?

Answer

The angle of reflection is the angle between the reflected light ray and a line perpendicular to the interface between the two media. The angle of reflection must be complementary to 20o.

90o – 20o = 70⁰

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Question

An incandescent light bulb is shown through a glass prism. The certain wavlength of the light is then directed into a glass cuvette containing an unknown concentration of protein. Commonly, this process is called spectroscopy and is used to determine the concentrations of DNA, RNA, and proteins in solutions. The indices of reflection of air, glass, and the solution are 1, 1.5, and 1.3, respectively.

The process of light entering glass from the air is called __________.

Answer

This question is asking us about the different processes that can happen to light upon hitting a surface. Thinking back to the properties of light, we know that absorption, reflection, and refraction are all processes that light can undergo when interacting with a surface.

Absorption means that the energy associated with the light is captured, and no photons are ejected from the surface after the collision of the incident photon and the surface. Reflection occurs when no light enters the new medium and instead bounces off at the angle to normal that it hit. Refraction occurs when some light enters the new medium. In this case, light is entering glass from the air; thus, the process we are concerned about is refraction.

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Question

Sound projected from an opera stage strikes a flat wall in the opera house at an angle $\small 37^o$ to the normal. What conclusion can be drawn about the reflection of this sound from the wall back into the room?

Answer

For all waves, the angle of incidence is equal to the angle of reflection. The projection is to the opposite side of the normal at the same angle as the incident wave.

Think of the wall as a mirror. The angle with which the wave impacts a mirror will be equal to the angle with which it is reflected, but mirrored across the normal.

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Question

Glass has an index of refraction of roughly . How much time would it take for light to pass through this glass if it were thick?

Answer

Since the index of refraction is 1.5, we can determine the speed of light in the glass using the following equation:

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

Rearranged to solve for velocity:

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

$v =\frac{3\cdot10^{8}\frac{m}{s}}{1.5}= 2\cdot10^{8}\frac{m}{s}$

Once the speed of light in the glass is known, we can use this quantity to determine how long it will take for the light to travel the width of the glass $\left(2cm)$ .

$t = \frac{0.02m}{2\cdot10^{^{8}}\frac{m}{s}}= 1\cdot10^{-10}sec$

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Question

A monochromatic light ray passes from air (n = 1.00) into glass (n = 1.50) at an angle of 20o with respect to the normal. What is the approximate angle of refraction?

Answer

To compare angles of incidence and refraction, use Snell's law.

$n_{1}sin(\theta {1})=n{2}sin(\theta _{2})$

$1.00sin(20)=1.50sin(\theta _{2})$

$.228=sin(\theta _{2})$

$\theta _{2}=sin^{-1}(.228)=13.2^o$

Notice that as the light enters a more dense medium, it bends towards the normal.

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Question

An incandescent light bulb is shown through a glass prism. The certain wavlength of the light is then directed into a glass cuvette containing an unknown concentration of protein. Commonly, this process is called spectroscopy and is used to determine the concentrations of DNA, RNA, and proteins in solutions. The indices of reflection of air, glass, and the solution are 1, 1.5, and 1.3, respectively.

The red light, selected for by the prism, is shown through air onto the glass cuvette at an angle of 30o to the normal. At what angle to normal does the light have when it is in the glass?

Answer

This problem asks us to consider refraction, that is, the bending of light when it enters a new medium. Thinking back to our light formulas in physics, we know that $n_1\sin\theta_1 = n_2\sin\theta_2$ , where n is the index of refraction of the medium and $\theta$ is the angle the light ray makes to normal.

Using the information provided in the pre-question text and the question above, we can solve for $\theta_2$ .

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

$sin\theta_2 = \frac{n_1\sin\theta_1}{n_2}$

$\theta_2 = \sin^{-1}(\frac{n_1\sin\theta_1}{n_2}) = \sin^{-1}(\frac{1*sin (30^o)}{1.5}) = 19.5^{\circ}$

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Question

An incandescent light bulb is shown through a glass prism. The certain wavlength of the light is then directed into a glass cuvette containing an unknown concentration of protein. Commonly, this process is called spectroscopy and is used to determine the concentrations of DNA, RNA, and proteins in solutions. The indices of reflection of air, glass, and the solution are 1, 1.5, and 1.3, respectively.

At what angle would the light passing through the glass cuvette need to hit the solution inside the cuvette for no light to enter the solution?

Answer

First, we need to determine what this question is asking us to do. If no light entered the solution from the glass, we know this is total internal reflection. Remember that total internal reflection occurs when light from one medium hits a second medium at an angle higher that the critical angle. Thinking back to our physics formulas, we know that the critical angle can be determined by the equation below.

$\theta_c = \sin^{-1}(\frac{n_2}{n_1})$

Plugging in the values for glass and the solution, we can find the critical angle.

$\theta_c = \sin^{-1}(\frac{n_2}{n_1}) = \sin^{-1}(\frac{1.3}{1.5}) = \sin^{-1}(0.87)=60.1^{\circ}$

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Question

An incandescent light bulb is shown through a glass prism. The certain wavlength of the light is then directed into a glass cuvette containing an unknown concentration of protein. Commonly, this process is called spectroscopy and is used to determine the concentrations of DNA, RNA, and proteins in solutions. The indices of reflection of air, glass, and the solution are 1, 1.5, and 1.3, respectively.

Red light is selected for by the prism and shown onto the glass cuvette at an angle of 30o to the normal. At what angle to normal does the light travel when it is in the solution after it has moved through wall of the cuvette?

Answer

Notice that, in this problem, there are two different changes in indices of refraction: air to glass and glass to solution; thus, we need to compute the angle change twice. First, let’s look at how the angle changes from air to glass. We know from Snell’s law that $n_1\sin\theta_1 = n_2\sin\theta_2$ , where n is the index of refraction of the medium and $\theta$ is the angle the light ray makes to normal.

Rearranging, we can find the angle to normal in the glass.

$sin\theta_2 = \frac{n_1\sin\theta_1}{n_2}$

$\theta_2 = \sin^{-1}(\frac{n_1\sin\theta_1}{n_2}) = \sin^{-1}(\frac{1sin (30^o)}{1.5}) = 19.5^{\circ}$

Now, we can use this angle and repeat the above equation to find the angle that the light enters the solution.

$n_2\sin\theta_2 = n_3\sin\theta_3$

$sin\theta_3 = \frac{n_2\sin\theta_2}{n_3}$

$\theta_3 = \sin^{-1}(\frac{n_2\sin\theta_2}{n_3}) = \sin^{-1}(\frac{1.5sin (19.5^o)}{1.3}) = 22.7^{\circ}$

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Question

An incandescent light bulb is shown through a glass prism. The certain wavlength of the light is then directed into a glass cuvette containing an unknown concentration of protein. Commonly, this process is called spectroscopy and is used to determine the concentrations of DNA, RNA, and proteins in solutions. The indices of reflection of air, glass, and the solution are 1, 1.5, and 1.3, respectively.

If the solution had a higher protein concentration than expected, how would the angle of refraction change as light travels from the glass into the solution?

Answer

This question is asking us to consider what would happen if the solution were more concentrated, essentially, if there were more particles per unit volume (denser). If we think back to the definition of index of refraction, we know that it relates to the density of a medium. The denser the medium, the higher the index of refraction.

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

Looking at Snell’s law, we can see the relationship between index of refraction and the angle of refraction.

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

$sin\theta_2 = \frac{n_1*\sin\theta_1}{n_2}$

If the solution were more concentrated, n2 would increase, making the term on the right side of the equation smaller. The sin function of a smaller number gives a smaller angle; thus, as the concentration increases (and thus the index of refraction increases), the angle of refraction gets smaller.

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Question

A scuba diver wearing a head lamp is returning to the surface of the Pacific Ocean. What is the index of refraction of the ocean water if the smallest angle resulting in total internal reflection is 35 degrees from the vertical?

$n_{air}=1.00$

Answer

We can use Snell's law to calculate the index of refraction of the water:

$n_1sin(\theta_1) = n_2sin(\theta_2)$

Where the subscript 1 denotes water and subscript 2 denotes air.

At the first incidence of total internal reflection, the angle of refraction is 90 degrees. Therefore, the sine function becomes 1, giving us the formula:

$n_1sin(\theta_1) = n_2$

Rearranging for the index of water:

$n_1 = \frac{n_2}{sin(\theta_1)} = \frac{1}{sin(35)} = 1.74$

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Question

The refractive index of medium A is 1.2, while that of medium B is 1.36. Through which medium does light travel faster and at what speed does it travel?

The speed of light is $3.0*10^8\frac{m}{s}$ .

Answer

The refractive index of a medium (n) is equal to the speed of light (c) divided by the velocity of light through the medium (v).
$n = \frac{c}{v}$
Rearranging the equation allows us to see the relationship regarding v.
$v = \frac{c}{n}$
The lower the refractive index, the faster the velocity of light. Medium A has the smaller refractive index. Light will travel faster through medium A at a velocity equal to the speed of light divided by the refractive index.

$v = \frac{\left ( 3 * 10^{8} m/s\right )}{1.2} = 2.5 * 10^{8} m/s$

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Question

An incandescent light bulb is shown through a glass prism. The certain wavlength of the light is then directed into a glass cuvette containing an unknown concentration of protein. Commonly, this process is called spectroscopy and is used to determine the concentrations of DNA, RNA, and proteins in solutions. The indices of reflection of air, glass, and the solution are 1, 1.5, and 1.3, respectively.

The velocity of the light __________ when it moves from air to glass.

Answer

This question asks us to consider the relationship between velocity and index of refraction of a medium. If we think back to the definition of index of refraction, we know that it is defined by the ratio of the velocity of light in a vacuum and the velocity of light in some other medium.

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

n is the index of refraction, c is the speed of light in a vacuum, and v is the speed of light in the new medium.

We can see that n and v are inversely proportional, meaning that the higher the n, the lower the velocity. As the light moved from air (n =1) to glass (n = 1.5), the n increased, and thus the velocity must decrease because the speed of light in a vacuum is constant.

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Question

An incandescent light bulb is shown through a glass prism. The certain wavlength of the light is then directed into a glass cuvette containing an unknown concentration of protein. Commonly, this process is called spectroscopy and is used to determine the concentrations of DNA, RNA, and proteins in solutions. The indices of reflection of air, glass, and the solution are 1, 1.5, and 1.3, respectively.

As light exits from the wall of the cuvette into the solution, its wavelength __________.

Answer

This question asks us to find the relationship between the wavelength and index of refraction. We will need to know two equations to compute the relationship between the two.

First, we need to relate velocity and index of refraction. The definition of index of refraction allows us to relate the two.

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

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

We also know the relationship between velocity and wavelength.

$v = \lambdaf$

We can now set these formulas equal to each other to find the relationship between wavelength and index of refraction.

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

We can see that $\lambda$ and n are inversely related. If n decreases, the wavelength must increase. In our problem, light in moving from a higher index of refraction to a lower one, meaning the wavelength gets longer (increases).

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Question

$n_{mineral\ oil}=1.46$

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

How long will it take a photon to travel $\small 55m$ through mineral oil?

Answer

The index of refraction is equal to the speed of light in a medium divided by the speed of light in a vacuum.

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

We can find the time to travel a given distance by manipulating this equation and combining it with the equation for rate: $\small v=\frac{d}{t}$ .

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

Plug in the given values and solve for the velocity in the medium.

$v = \frac{310^8 \frac{m}{s}}{1.46} = 2.0510^8 \frac{m}{s}$

Now we can return to the rate equation and solve for the time to travel $\small 55m$ .

$v=\frac{d}{t}$

$2.0510^8\frac{m}{s}=\frac{55m}{t}$

$t=2.6810^{-7}s$

We can recognize that the answer can be simplified by converting to nanoseconds.

$2.6810^-7=26810^{-9}=268ns$

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Question

Which of the following does not take place when a light wave travels from a medium with a high index of refraction into one with a lower index of refraction?

Answer

As the wave travels into the less dense medium, it speeds up, bending away from the normal line. The index of refraction tells the ratio of the velocity in a vacuum in relation to the velocity the medium; thus, the velocity will be greater in a medium with a lower index of refraction.

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

Frequency remains the same regardless of medium, however, since the velocity changes, the wavelength must accommodate this change.

$v=f\lambda$

If velocity increases and frequency remains constant, wavelength must also increase.

Finally, a phase shift only occurs when a light ray reflects from the surface of a more dense medium.

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Question

What is the index of refraction for a material in which light travels at $1.85*10^{8}\frac{m}{s}$ ?

Answer

Relevant equations:

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

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

To find index of refraction, divide the speed of light in a vacuum by the speed of light in the material:

$n = \frac{3 * 10^8}{1.85 * 10^8} \approx 1.62$

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Question

The refractive index of medium A is 1.2, while that of medium B is 1.36. Through which medium does light travel faster and at what speed does it travel?

The speed of light is $3.0*10^8\frac{m}{s}$ .

Answer

The refractive index of a medium (n) is equal to the speed of light (c) divided by the velocity of light through the medium (v).
$n = \frac{c}{v}$
Rearranging the equation allows us to see the relationship regarding v.
$v = \frac{c}{n}$
The lower the refractive index, the faster the velocity of light. Medium A has the smaller refractive index. Light will travel faster through medium A at a velocity equal to the speed of light divided by the refractive index.

$v = \frac{\left ( 3 * 10^{8} m/s\right )}{1.2} = 2.5 * 10^{8} m/s$

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Question

A light ray traveling through a medium is reflected by a second medium at an angle of 20⁰ to the interface between the two media. Which of the following is true?

Answer

The angle of reflection is the angle between the reflected light ray and a line perpendicular to the interface between the two media. The angle of reflection must be complementary to 20o.

90o – 20o = 70⁰

Compare your answer with the correct one above

Question

An incandescent light bulb is shown through a glass prism. The certain wavlength of the light is then directed into a glass cuvette containing an unknown concentration of protein. Commonly, this process is called spectroscopy and is used to determine the concentrations of DNA, RNA, and proteins in solutions. The indices of reflection of air, glass, and the solution are 1, 1.5, and 1.3, respectively.

The process of light entering glass from the air is called __________.

Answer

This question is asking us about the different processes that can happen to light upon hitting a surface. Thinking back to the properties of light, we know that absorption, reflection, and refraction are all processes that light can undergo when interacting with a surface.

Absorption means that the energy associated with the light is captured, and no photons are ejected from the surface after the collision of the incident photon and the surface. Reflection occurs when no light enters the new medium and instead bounces off at the angle to normal that it hit. Refraction occurs when some light enters the new medium. In this case, light is entering glass from the air; thus, the process we are concerned about is refraction.

Compare your answer with the correct one above

Question

Sound projected from an opera stage strikes a flat wall in the opera house at an angle $\small 37^o$ to the normal. What conclusion can be drawn about the reflection of this sound from the wall back into the room?

Answer

For all waves, the angle of incidence is equal to the angle of reflection. The projection is to the opposite side of the normal at the same angle as the incident wave.

Think of the wall as a mirror. The angle with which the wave impacts a mirror will be equal to the angle with which it is reflected, but mirrored across the normal.

Compare your answer with the correct one above

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