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MCAT Physical : MCAT Physical Sciences

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8 Diagnostic Tests 303 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Reflection And Refraction

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?

Possible Answers:

The reflection will be $\small 74^o$ to the opposite side of the normal

The reflection will be $\small 53^o$ to the same side of the normal

The reflection will be $\small 53^o$ to the opposite side of the normal

The reflection will be $\small 37^o$ to the opposite side of the normal

The reflection will be $\small 37^o$ to the same side of the normal

Correct answer:

The reflection will be $\small 37^o$ to the opposite side of the normal

Explanation:

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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Example Question #2 : Reflection And Refraction

Glass has an index of refraction of roughly 1.5 . How much time would it take for light to pass through this glass if it were 2cm thick?

Possible Answers:

$6\cdot10^{6} s$

$4\cdot10^{-11} s$

$2\cdot10^{-10}s$

$1\cdot10^{^{-10}} s$

Correct answer:

$1\cdot10^{^{-10}} s$

Explanation:

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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Example Question #201 : Mcat Physical Sciences

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.

Possible Answers:

cannot be determined

increases

decreases

remains the same

Correct answer:

decreases

Explanation:

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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Example Question #21 : Optics

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

Possible Answers:

remains the same

decreases

increases

cannot be determined

Correct answer:

increases

Explanation:

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.

$v = \lambda*f$

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

$\lambda*f = \frac{c}{n}$

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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Example Question #1 : Velocity And Index Of Refraction

$n_{mineral\ oil}=1.46$

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

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

Possible Answers:

1.83ns

268ns

2.02ns

2.67ns

183ns

Correct answer:

268ns

Explanation:

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.

$v = \frac{3*10^8 \frac{m}{s}}{1.46} = 2.05*10^8 \frac{m}{s}$

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

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

$2.05*10^8\frac{m}{s}=\frac{55m}{t}$

$t=2.68*10^{-7}s$

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

$2.68*10^-7=268*10^{-9}=268ns$

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Example Question #21 : Optics

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 $3.0*10^8\frac{m}{s}$ .

Possible Answers:

Medium B at velocity $2.8*10^8\frac{m}{s}$

The speed of light will be $3.0*10^8\frac{m}{s}$ in either medium

Medium A at velocity $1.8*10^8\frac{m}{s}$

Medium B at velocity $2.2*10^8\frac{m}{s}$

Medium A at velocity $2.5*10^8\frac{m}{s}$

Correct answer:

Medium A at velocity $2.5*10^8\frac{m}{s}$

Explanation:

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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Example Question #171 : Ap Physics 2

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?

Possible Answers:

The refracted light remains in phase with the incident wave

Velocity increases

Wavelength increases

Frequency increases

The light ray bends away from the normal line

Correct answer:

Frequency increases

Explanation:

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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Example Question #2 : Velocity And Index Of Refraction

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

Possible Answers:

Such a material does not exist, since the speed of light is constant

2.61

0.54

1.85

1.62

Correct answer:

1.62

Explanation:

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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Example Question #1 : Snell's Law

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

Possible Answers:

17^o

31^o

24^o

27^o

13^o

Correct answer:

13^o

Explanation:

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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Example Question #204 : Mcat Physical Sciences

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

Possible Answers:

19.5^o

45^o

22^o

30^o

Correct answer:

19.5^o

Explanation:

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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All MCAT Physical Resources

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MCAT Physical : MCAT Physical Sciences

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #1 : Reflection And Refraction

Example Question #2 : Reflection And Refraction

Example Question #201 : Mcat Physical Sciences

Example Question #21 : Optics

Example Question #1 : Velocity And Index Of Refraction

Example Question #21 : Optics

Example Question #171 : Ap Physics 2

Example Question #2 : Velocity And Index Of Refraction

Example Question #1 : Snell's Law

Example Question #204 : Mcat Physical Sciences

All MCAT Physical Resources

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