Types of Image - MCAT Physical
Card 0 of 21
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?
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?
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 
is equal to 
. Since 
is a positive number we know the image is real, and thus inverted.
The second equation to consider is for magnification: 
.
If the absolute value of 
is greater than one, the image is magnified, and if the value is less than one, it is minimized.
We would expect the image to be 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.
Compare your answer with the correct one above
For a nearsighted person, the image of a distant object is formed __________.
For a nearsighted person, the image of a distant object is formed __________.
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.
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.
Compare your answer with the correct one above
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?
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?
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 
, then the image is inverted, real, and smaller than the object.
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
Compare your answer with the correct one above
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?
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?
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 is equal to . Since is a positive number we know the image is real, and thus inverted.
The second equation to consider is for magnification: .
If the absolute value of is greater than one, the image is magnified, and if the value is less than one, it is minimized.
We would expect the image to be 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.
Compare your answer with the correct one above
For a nearsighted person, the image of a distant object is formed __________.
For a nearsighted person, the image of a distant object is formed __________.
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.
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.
Compare your answer with the correct one above
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?
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?
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 , then the image is inverted, real, and smaller than the object.
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
Compare your answer with the correct one above
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?
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?
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 is equal to . Since is a positive number we know the image is real, and thus inverted.
The second equation to consider is for magnification: .
If the absolute value of is greater than one, the image is magnified, and if the value is less than one, it is minimized.
We would expect the image to be 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.
Compare your answer with the correct one above
For a nearsighted person, the image of a distant object is formed __________.
For a nearsighted person, the image of a distant object is formed __________.
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.
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.
Compare your answer with the correct one above
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?
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?
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 , then the image is inverted, real, and smaller than the object.
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
Compare your answer with the correct one above
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?
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?
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 is equal to . Since is a positive number we know the image is real, and thus inverted.
The second equation to consider is for magnification: .
If the absolute value of is greater than one, the image is magnified, and if the value is less than one, it is minimized.
We would expect the image to be 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.
Compare your answer with the correct one above
For a nearsighted person, the image of a distant object is formed __________.
For a nearsighted person, the image of a distant object is formed __________.
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.
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.
Compare your answer with the correct one above
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?
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?
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 , then the image is inverted, real, and smaller than the object.
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
Compare your answer with the correct one above
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?
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?
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 is equal to . Since is a positive number we know the image is real, and thus inverted.
The second equation to consider is for magnification: .
If the absolute value of is greater than one, the image is magnified, and if the value is less than one, it is minimized.
We would expect the image to be 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.
Compare your answer with the correct one above
For a nearsighted person, the image of a distant object is formed __________.
For a nearsighted person, the image of a distant object is formed __________.
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.
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.
Compare your answer with the correct one above
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?
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?
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 , then the image is inverted, real, and smaller than the object.
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
Compare your answer with the correct one above
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?
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?
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 is equal to . Since is a positive number we know the image is real, and thus inverted.
The second equation to consider is for magnification: .
If the absolute value of is greater than one, the image is magnified, and if the value is less than one, it is minimized.
We would expect the image to be 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.
Compare your answer with the correct one above
For a nearsighted person, the image of a distant object is formed __________.
For a nearsighted person, the image of a distant object is formed __________.
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.
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.
Compare your answer with the correct one above
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?
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?
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 , then the image is inverted, real, and smaller than the object.
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
Compare your answer with the correct one above
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?
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?
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 is equal to . Since is a positive number we know the image is real, and thus inverted.
The second equation to consider is for magnification: .
If the absolute value of is greater than one, the image is magnified, and if the value is less than one, it is minimized.
We would expect the image to be 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.
Compare your answer with the correct one above
For a nearsighted person, the image of a distant object is formed __________.
For a nearsighted person, the image of a distant object is formed __________.
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.
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.
Compare your answer with the correct one above