MCAT Physical : MCAT Physical Sciences
Study concepts, example questions & explanations for MCAT Physical
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Example Questions
Example Question #122 : Ap Physics 2
A person approaches a plane mirror at 5m/s. How fast do they approach the mirror image?
5m/s
10m/s
7.5m/s
2.5m/s
10m/s
The image distance for a plane mirror is always equal to the object distance because the magnification is 1.
If the object and image are the same distance from the mirror and magnification is 1, then as the object approaches the mirror at a certain speed, the image is approaching the plane mirror at the same speed, therefore you approach the image more quickly than you approach the mirror, since you travel 5m/s toward the mirror and the image travels 5m/s toward the mirror.
Example Question #1 : Mirrors And Lenses
In a microscope has a 
The total magnification of a compound microscope is the product of the objective lens magnification and the eyepiece magnification.
Objective magnification and eyepiece magnification are given by the following equations:
We are given the tube length and focal length of the objective lens, allowing us to solve for its magnification.
We also know the distance of the viewer's eye. Use this value in the eyepiece magnification equation.
Finally, combine the eyepiece magnification and objective lens magnification into the original equation for total magnification.
Use the given value for total magnification to solve for the focal length of the eyepiece lens.
Example Question #11 : Mirrors And Lenses
A compound microscope consists of an eyepiece with an angular magnification of 25 and an objective lens of unknown focal length. If the length of the microscope tube is 25cm, what magnitude of objective focal length is necessary to achieve an overall magnification of 500?
Relevant equations:
Given:
Step 1: Plug the expression for 
Step 2: Rearrange to isolate the unknown,
Step 3: Plug in given quantities, taking the absolute value to find the magnitude of 
Example Question #1 : Combining Lenses
What diameter of telescope objective lens is necessary to resolve two stars that primarily emit 600nm light and have an angular separation of 
Relevant equations:
Step 1: Rearrange equation to isolate the unknown, 
Step 2: Plug in the given numbers for wavelength and angular separation:
Example Question #2 : Gre Subject Test: Physics
A man stands ten meters away from a converging mirror with a focal length of two meters. What is true of the image he sees?
His image is real, inverted, and minimized
His image is virtual, upright, and minimized
His image is virtual, upright, and magnified
His image is real, upright, and magnified
His image is real, inverted, and magnified
His image is real, inverted, and minimized
The first thing to consider when answering this question is the fact that real images are always inverted and virtual images are always upright. Once you have determined one or the other, two answer choices can be eliminated.
The first equation that is necessary for this question is 
From this we can determine that 
The second equation to consider is for magnification: 
If the absolute value of 
We would expect the image to be minimized.
Example Question #1 : Types Of Image
For a nearsighted person, the image of a distant object is formed __________.
behind the retina, and corrected using a diverging lens
in front of the retina, and corrected using a diverging lens
behind the retina, and corrected using a converging lens
in front of the retina, and corrected using a converging lens
in front of the retina, and corrected using a diverging lens
In nearsightedness, the person cannot see far objects due to increased refraction. This causes the image to be formed in front of the retina. This condition is corrected using a diverging lens to compensate for the "over refraction" by the deformed cornea.
Example Question #1 : Types Of Image
An object is placed 50cm in front of a converging lens whose focal length is 20cm. Which of the following best describes the image that is formed?
Larger than the object and virtual
The same size as the object and virtual
Larger than the object and real
Smaller than the object and virtual
Smaller than the object and real
Smaller than the object and real
Relevant equations:
Step 1: Plug in the given focal length and object distance to find the image distance:
Step 2: Find the magnification and orientation of the image:
A negative magnification means the image is inverted, and therefore real. A magnification smaller than 1 means the image is smaller than the object.
The following result is true, in general, for converging lenses: if the object is outside 
Example Question #121 : Ap Physics 2
Jimmy is farsighted and uses a convex lens to correct his vision. Wendy is nearsighted and uses a concave lens to correct her vision. They both wear glasses. During a camping trip, they notice they do not have any matches, and decide to use their glasses to start the fire. Whose glasses could be used to start the fire?
Jimmy
Both glasses would work equally well
Neither glasses would work
Wendy
Jimmy
This question deals with an application of optics. In this case we have a farsighted person and a near sighted person. The farsighted person would use a convex lens, which is a converging lens. This would allow all of the rays of light to converge on a single point, allowing them to heat the object up and start a fire. Wendy’s glasses are diverging lenses, which would cause the rays to separate.
Example Question #1 : Reflection And Refraction
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?
The refractive indices of the media must be equal
The refractive index of the second medium is equal to sin(20⁰)
The angle of reflection is 70o
The frequency of the light wave is greater upon reflection
The angle of reflection is 70o
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
Example Question #2 : General Principles And Properties
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 __________.
reflection
diffraction
absorption
refraction
refraction
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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