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Test: Spherical Mirrors - MCAT MCQ

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10 Questions MCQ Test - Test: Spherical Mirrors

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Test: Spherical Mirrors - Question 1

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Which of the following correctly describes the image produced by a plane mirror?

A.
The image is real and upright.
B.
The image will move twice as fast towards the plane of the mirror as the object moves.
C.
The image is the same size as the object with no lateral inversion.
D.
The image undergoes vertical inversion.

Detailed Solution for Test: Spherical Mirrors - Question 1

The image is virtual and upright since the image appears where the light rays project back upon reflection.
The image is not vertically, but it is inverted laterally.
The image is the same size as the object, but with lateral inversion.
The image will appear to move twice as fast as the object. As instance, if a person is standing 2 meters from the mirror, then the image appears to be 4 meters from the person. For the two to come together, the person travels 2 meters, while the image appears to travel 4 meters in the same time period.

Test: Spherical Mirrors - Question 2

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A student who is standing 0.5 meters from a plane mirror moves at a velocity of 0.75 meters per second for 3 seconds away from the mirror. How much further is the student from her image than when she first started moving?

A.
5 meters
B.
3¼ meters
C.
2¼ meters
D.
4½ meters

Detailed Solution for Test: Spherical Mirrors - Question 2

The student is standing 0.5 meters from the plane mirror, so she is starting 1 meter from her own image.
The question stem states the student then moves 0.75 m/s for 3 seconds for a total distance of 2.25 meters. At the end, she is 2.75 meters from the plane mirror.
Since the plane mirror creates an equidistant virtual image, she must be 5.50 meters from her own image.
The student is ultimately 4.50 or 4½ meters further from her image than when she first started moving.

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Test: Spherical Mirrors - Question 3

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Which of the following statements most accurately describes virtual image formation?

A.
An enlarged, virtual, and inverted image can be formed by a positive lens when the object is placed inside the focal point.
B.
A virtual image is formed at the position where the paths of the principal rays cross when projected backward from their paths beyond the lens.
C.
A virtual image does not form a visible projection on a screen and cannot be seen by most optical instruments except for the human eye.
D.
A virtual, upright, and reduced image is formed by a single positive lens regardless of the object position.

Detailed Solution for Test: Spherical Mirrors - Question 3

Let’s review some characteristics of virtual images as opposed to real images. While real images can form a visible projection on a screen, virtual images can only be imaged by optical instruments like the eye, camera, or other optical instruments.
A virtual, upright, and reduced image only can be formed by a negative lens regardless of the object position.
Meanwhile, a virtual, upright, and enlarged image can be formed by a positive lens only when the object is placed inside the focal point.
A virtual image is formed at the position where the paths of the principal rays cross when projected backed from their paths.

Test: Spherical Mirrors - Question 4

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Which of the following images can be formed by a concave mirror?
I. real, inverted, and enlarged
II. virtual, upright, and enlarged
III. real, upright, and reduced
IV. virtual, inverted, and enlarged

A.
I and II
B.
I and III
C.
I, III, and IV
D.
II and IV

Detailed Solution for Test: Spherical Mirrors - Question 4

This can be solved by using ray diagrams or through calculation. Let’s achieve this through calculation with a concave mirror with a focal length of 5 cm.
While the equation is 1/f = 1/i + 1/0, let’s use this rearrangement of the formula:
In the case of an image outside the radius of curvature, using 20 cm as the object distance:
In the case of an image within the radius of curvature but outside the focal point, using 8 cm as the object distance:

The image produced is real (positive i), inverted (negative m), and enlarged (m > 1).
In the case of an image within the focal point, using 3 cm as the object distance

The image produced is virtual (negative i) , upright (positive m), and enlarged (m > 1).

Test: Spherical Mirrors - Question 5

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We have our first experience with rearview mirrors when first learning to drive. Which of the following statements most accurately describes rearview mirrors?

A.
Rearview mirrors are converging such that images appear closer than their actual distance.
B.
Rearview mirrors are diverging such that the image produced is virtual, upright, and enlarged.
C.
Rearview mirrors are converging such that the image produced is real, upright, and reduced
D.
Rearview mirrors are diverging such that the object distance is greater than the image distance.

Detailed Solution for Test: Spherical Mirrors - Question 5

When we look into rearview mirrors, the image appears on the other side of the mirror. We can deduce that it must be a virtual image.
Converging mirrors produce a virtual and enlarged image, while diverging mirrors produce a virtual and reduced image.
A ray diagram must be drawn to determine the image distance compared with the object distance. Only for diverging mirrors is the object distance greater than the image distance.
As a final note, there is a warning that objects are closer than they appear. Although the virtual image is closer than the actual object, ct. the angular size of the virtual image is smaller than the angular size of the object. We tend to estimate distance by angular size rather than by the angle of convergence of our eyes, so the think the object is farther than it really is.

Test: Spherical Mirrors - Question 6

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A car drives by approximately 20 meters from the convex side of a reflective sculpture, and its image appears to be one-fifth the size of the car. What is the focal length of this reflective surface?

A.
3.3 meters
B.
-3.3 meters
C.
0.2 meters
D.
-5.0 meters

Detailed Solution for Test: Spherical Mirrors - Question 6

Let’s determine what is given in the question stem. The object distance is 20 meters, and the magnification is +⅕ since we can assume that the image is upright.
We use the magnification formula to determine the image distance:
Next we use the mirror equation to determine the focal length:
The focal length is -5.0 meters, and it is negative because there is a diverging or convex mirror.

Test: Spherical Mirrors - Question 7

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Which of the following would be an example of using a convex mirror effectively?

A.
Satellite dishes for collecting radio waves
B.
Otoscopic mirror for concentrating light
C.
Dental mirror for closer examination
D.
Security mirror in store to increase range of view

Detailed Solution for Test: Spherical Mirrors - Question 7

A convex mirror is diverging, which would make light disperse if you shine light upon its surface. A converging mirror would be more useful for collecting light onto a smaller point or area like in telescope.
Therefore, it would not be useful for in satellite dishes or in an otoscopic mirror.
A convex mirror also produces an image that is smaller than the actual object. Therefore, it would not be useful in a dental mirror for closer examination.
However, it is useful in increasing the range of view for security mirrors in stores. A result of that increased range is the small size of all objects.

Test: Spherical Mirrors - Question 8

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Which of the following statements correctly identifies a rule for drawing ray diagrams with mirrors?

A.
A ray incident along the radius vector is reflected back on itself.
B.
A light ray parallel to the principal axis of the mirror is reflected so that it appears to come from the radius of curvature.
C.
A ray that strikes the mirror at its center at the Brewster angle is reflected symmetrically backwards by the same angle below the axis.
D.
A light ray pointed directly at the focal point is reflected such that the angle of incidence is equal to the angle of reflection.

Detailed Solution for Test: Spherical Mirrors - Question 8

An incident ray parallel to the principal axis will pass through the focal point upon reflection for converging or is reflected so that it appears to come from the focal point for diverging.
An incident ray passing through the focal point will travel parallel to the principal axis upon reflection.
A ray that strikes the mirror at its center will be reflected symmetrically backwards by the same angle below the axis. Brewster angle is the angle at which reflected light becomes all polarized.
A ray incident along the radius vector will be reflected back on itself.

Test: Spherical Mirrors - Question 9

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A clown is standing in front of a fun mirror. The top half of the mirror is concave, and the bottom half of the mirror is convex. What happens to his image as the clown moves towards the mirror?

A.
His top half gets bigger up to his actual size; his bottom half gets smaller down to his actual size.
B.
His top half gets smaller down to his actual size; his bottom half gets bigger up to his actual size.
C.
His top half gets smaller beyond his actual size; his bottom half gets bigger beyond his actual size.
D.
His top half gets bigger beyond his actual size; his bottom half stays the same size.

Detailed Solution for Test: Spherical Mirrors - Question 9

For a concave mirror, we have determined that there are 3 images that can be formed, but the order must be determined: outside radius of curvature to within the radius of curvature to within the focal point.
As the object moves towards the focal point, the real and inverted image starts smaller than the object beyond the radius of curvature. The image gets bigger and bigger, and once past the radius of curvature, it is now larger than the object.
At the focal point, the image is at infinity. As the object moves past the focal point, the virtual and upright image that is infinitely large gets smaller and smaller. The image does not get smaller than the object itself.
As for a diverging mirror, there is usually one image that can be formed. Let’s say the object distance is 3 cm and the focal length is 5 cm, remembering that it will be negative for diverging mirrors:

The image is virtual (negative i), reduced (m<1), and upright (positive m).
For diverging mirrors, the image gets larger and larger as the object moves towards the lens. The image never gets larger than the object.

Test: Spherical Mirrors - Question 10

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Which of the following statements best describes the causes and solutions for optical system aberrations?

A.
Spherical aberration occurs most at the center of a mirror such that the reflected rays deviate the most from the center as compared to the edges.
B.
Chromatic aberration is commonly corrected using an additional lens to ensure that all the wavelengths converge at the focal point.
C.
Chromatic aberration occurs with converging lens because different wavelengths converge at different points.
D.
Spherical aberration only occurs with mirrors and chromatic aberration with lens.

Detailed Solution for Test: Spherical Mirrors - Question 10

Spherical aberration is a phenomenon observed in either a lens or mirror due to the increased refraction or reflection of light rays near its edge as compared to those that strike its center.
Chromatic aberration is a phenomenon observed in lens in which there is a failure of different wavelengths of light to converge at the same point.
Spherical aberration can be most easily corrected by using a parabolic shaped mirror or specially shaped or doublet lens.
Chromatic aberration can be most easily corrected by the use of glasses of different dispersion either in a singlet or a doublet.

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