Test: Thin Lenses - MCAT MCQ

A converging lens can produce 3 different categories of images: 1) real, inverted, enlarged; 2) real, inverted, reduced; and 3) virtual, upright, enlarged.

Since the fourth image type, which is virtual, upright, and reduced, is only produced by a diverging lens. Since a virtual, upright, and enlarged image was produced in its final state, we can deduce that the lens was not switched from a converging to a diverging lens.

By adding an additional converging lens, it is assumed that the original lens is also converging. Two converging lens will produce a real and upright image since the first produces a real, inverted, and reduced image and the second will invert it again.

The image is real, inverted, and reduced when the object is beyond the radius of curvature.

The image is real, inverted, and enlarged when the object is within the radius of curvature and beyond the focal point.

The image is virtual, upright, and enlarged when the object is within the focal point. Therefore, the correct answer is that the object is moved from a point outside the focal length to a point within the focal length for a convex or converging lens.

• For a converging lens, the image is real, inverted, and reduced when the object is positioned at greater than twice the focal length or radius of curvature (O > 2F), real, inverted, and enlarged when the object is between the radius of curvature and the focal length (2F > O > F), and virtual, upright, and enlarged when the object is positioned within the focal length (O < F).
• The image is always real and inverted not within but outside the focal length of the lens.
• The image is always virtual and upright not outside the radius of curvature but within the focal point of the lens.
• The image is always smaller than the object outside not the focal length but the radius of curvature of the lens.
• Correctly stated, the image is always larger than the object within the radius of curvature of the lens.

• The concave or diverging lens can only produce one type of image. This can be determined by a number of methods: ray diagram, thin lens equation, or memorization.
• In drawing a ray diagram, there are 3 to draw: the ray parallel to the axis and through the focal point on the opposite side, the ray through the focal point and parallel to the axis on the opposite side, and the ray that passes through the center unchanged.
• In the diagram below, the ray diagram indicates that a virtual, upright, and reduced image is produced:
  

• The refractive power of a lens that makes light rays more convergent is positive. The focal length of such a lens is positive.
• Generally, longitudinal distances are positive if pointing to the right.
• Light propagates from left to right such that the distance of a real image is positive since it appears where light propagates and the distance of a virtual image is negative since it appears on the same side of the lens.
• Lastly, the magnitude does indicate whether an image is reduced or enlarged, and the signage indicates whether it is inverted or upright.

• This is a straightforward application of the thin lens formula although you could get a rough estimate from drawing a ray diagram:
  
• The values from the question stem are f = -15 cm and o = 45 cm. The focal length for diverging lens are negative. Object length is positive moving left starting from the lens.
• Plug in your values:
  
• Magnification is calculated from the image and object distances:
  
• The image is 11.25 cm to the left of the lens with a reduced size of 0.25.

The equation that should be used to solve this problem is:

The height of the image on the retina is -0.12 centimeters, which tells us also that the image is inverted.

• This problem can be solved with some ray diagrams with some deduction or with the thin lens equation performed twice.
• The first lens is diverging lens since the focal length is negative, and the second lens is converging since the focal length is positive.
• The object is placed well outside the focal length for the diverging lens, but regardless of position, the image produced is virtual, upright, and reduced. The further away from the focal length, the smaller the image.
• This virtual image is the object for the converging lens, which falls well outside the radius of curvature or twice the focal length. In this case, the second lens will produce a real, inverted, and reduced image.

• The telescope is equal to the sum of the focal lengths of the objective and eyepiece. The focal length should have positive values they are both converging.
• Since the objective lens is responsible for gathering light.
• The telescope makes use of two converging lenses. The objective produces a real and inverted image. If you follow the light rays, they switch sides when refracting through the lens.
• The eyepiece acts like a simple magnifier to enlarge the image formed by the objective. That image should fall on or within the focal point of the eyepiece.
• Since the image from the objective is already inverted, the image produced by the eyepiece will be virtual, enlarged, and inverted.

• The compound microscope uses an objective lens that has a short focal length, and the object sits outside the focal length of the objective.
• The objective is close to the object being viewed which focuses a real and inverted image of the object inside the microscope. The objective has a short focal length to produce an image that is roughly the same size and orientation as the object.
• The eyepiece then acts a simple magnifier since the image from the objective falls within the focal length of the eyepiece. When the object falls within the focal length, the eyepiece gives an virtual, upright, and enlarged image of the object.
• However, the image is already inverted, so here the final image is virtual, inverted, and enlarged. So the microscope produces an image similar to the telescope, which is also a virtual, inverted, and enlarged image.
• By decreasing the focal length of the objective and increasing the diameter of the objective for a microscope will increase the magnifying and light-gathering power respectively. Magnification is the power of the objective multiplied by the power of the lens, and power is inversely related to focal length.
• The correct answer is the false statement. The image from the objective lens is produced within the focal point of the eyepiece, which then produces a real, upright, and enlarged image on the retina.

• Changing the objective to a diverging lens would not properly gather light into the telescope in producing a virtual image outside the telescope. Only a converging lens would work for the first lens.
• Changing the eyepiece to a converging lens would not invert the image vertically but laterally. The virtual image would be projected back into the telescope rather than on the retina.
• Moving the eyepiece to within the focal point of the objective such that they overlap would require the additional step of changing the eyepiece to a diverging lens to create the virtual, upright, and enlarged image:
  
• Adding a third converging lens in between such that the image from the objective falls 2 focal lengths from the lens would properly invert the image: