MCAT Physical › Mirrors and Lenses
A concave mirror has a radius of curvature of 0.85m. Where is the mirror's focal point?
The focal point cannot be found from the given information
Since the mirror is concave, the focal point will be in front of the mirror. The focal length is equal to one half of the radius of curvature.
Rc is the radius of curvature. Plugging in 0.85m for Rc allows us to solve for the focal length.
0.43m is equal to 43cm.
A concave mirror has a radius of curvature of 0.85m. Where is the mirror's focal point?
The focal point cannot be found from the given information
Since the mirror is concave, the focal point will be in front of the mirror. The focal length is equal to one half of the radius of curvature.
Rc is the radius of curvature. Plugging in 0.85m for Rc allows us to solve for the focal length.
0.43m is equal to 43cm.
An object is placed 50cm in front of a concave mirror of radius 60cm. How far from the mirror is the image?
Relevant equations:
Step 1: Find the focal length of the mirror.
Step 2: Plug this focal length and the object distance into the lens equation.
An object is placed 50cm in front of a concave mirror of radius 60cm. How far from the mirror is the image?
Relevant equations:
Step 1: Find the focal length of the mirror.
Step 2: Plug this focal length and the object distance into the lens equation.
How far from a converging lens must an object be placed to produce an image that is NOT real and inverted? Given the answer as in terms of the focal length,
.
When an object is placed a distance from a converging lens or mirror that is equal to the focal length, no image is produced. To test this out, stand in front of a single concave mirror and continue to back up until you no longer see an image. Once you've reached this point, you will be standing one focal length away from the mirror.
When the object is less than one focal length away from the converging lens/mirror, the image will be virtual and upright.
If you don't have these trends committed to memory, you can derive them from the equation .
When is a negative integer the image is virtual, and when it is a positive integer the image is real.
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 real, inverted, and magnified
His image is virtual, upright, and magnified
His image is virtual, upright, and minimized
His image is real, upright, and magnified
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.
An object that is 3cm tall is placed 30cm from a convex spherical mirror of radius 40cm, along its central axis. What is the height of the image that is formed?
Relevant equations:
Step 1: Find the focal length of the mirror (remembering that convex mirrors have negative focal lengths, by convention).
Step 2: Find the image distance using the thin lens equation.
Step 3: Use the magnification equation to relate the object distances and heights.
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 real, inverted, and magnified
His image is virtual, upright, and magnified
His image is virtual, upright, and minimized
His image is real, upright, and magnified
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.
How far from a converging lens must an object be placed to produce an image that is NOT real and inverted? Given the answer as in terms of the focal length,
.
When an object is placed a distance from a converging lens or mirror that is equal to the focal length, no image is produced. To test this out, stand in front of a single concave mirror and continue to back up until you no longer see an image. Once you've reached this point, you will be standing one focal length away from the mirror.
When the object is less than one focal length away from the converging lens/mirror, the image will be virtual and upright.
If you don't have these trends committed to memory, you can derive them from the equation .
When is a negative integer the image is virtual, and when it is a positive integer the image is real.
An object that is 3cm tall is placed 30cm from a convex spherical mirror of radius 40cm, along its central axis. What is the height of the image that is formed?
Relevant equations:
Step 1: Find the focal length of the mirror (remembering that convex mirrors have negative focal lengths, by convention).
Step 2: Find the image distance using the thin lens equation.
Step 3: Use the magnification equation to relate the object distances and heights.