Types of Image - MCAT Chemical and Physical Foundations of Biological Systems

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.