Learning Goal: To learn the quantitative use of the lens equation, as well as ho

Question

Learning Goal: To learn the quantitative use of the lens equation, as well as how to determine qualitative properties of solutions.

In working with lenses, there are three important quantities to consider: The object distance s is the distance along the axis of the lens to the object. The image distance s' is the distance along the axis of the lens to the image. The focal length f is an intrinsic property of the lens. These three quantities are related through the equation

1/s + 1/s' = 1/f.

Note that this equation is valid only for thin, spherical lenses. Unless otherwise specified, a lens problem always assumes that you are using thin, spherical lenses.

The equation above allows you to calculate the locations of images and objects. Frequently, you will also be interested in the size of the image or object, particularly if you are considering a magnifying glass or microscope. The ratio of the size of an image to the size of the object is called the magnification. It is given by

m = y'/y = -s'/s,

where y' is the height of the image and y is the height of the object. The second equality allows you to find the size of the image (or object) with the information provided by the thin lens equation.

All of the quantities in the above equations can take both positive and negative values. Positive distances correspond to real images or objects, while negative distances correspond to virtual images or objects. Positive heights correspond to upright images or objects, while negative heights correspond to inverted images or objects. The following table summarizes these properties:
Positive Negative
s real virtual
s' real virtual
y upright inverted
y' upright inverted

The focal length f can also be positive or negative. A positive focal length corresponds to a converging lens, while a negative focal length corresponds to a diverging lens.

Consider an object with s = 12 cm that produces an image with s' = 15 cm. Note that whenever you are working with a physical object, the object distance will be positive (in multiple optics setups, you will encounter "objects" that are actually images, but that is not a possibility in this problem). A positive image distance means that the image is formed on the side of the lens from which the light emerges.
Now consider a diverging lens with focal length f=-15 cm, producing an upright image that is 5/9 as tall as the object.
F.) What is the object distance? You will need to use the magnification equation to find a relationship between s and s'. Then substitute into the thin lens equation to solve for s.


A lens placed at the origin with its axis pointing along the x axis produces a real inverted image at x=-24 cm that is twice as tall as the object.
H.) What is the image distance?

Explanation / Answer

F) The focal length of the diverging lens ia f = -15 cm The image has size h' = 5h/9 Where h is the height of the object The image is upright means that the magnification is positive the magnification m = -v/u v is the image distance and u is the object distance m = -v/u = h'/h -v/u = 5/9 The equation for the lens is 1/f = 1/u + 1/v u/f = u/v + 1 u/-15 = -9/5 + 1 u = -4*-15/5 u = 12 cm The object is at distance 12 cm from the lens H) The image distance is v = -24 cm

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