A diverging lens has a focal length of -8.00 cm. (a) What are the image distances for objects p

step 1 Here we're going to look at the imaging by a diverging lens. A diverging lens has a negative foc

A diverging lens has a focal length of -8.00 cm. (a) What...

A diverging lens has a focal length of $-8.00 \mathrm{~cm}$. (a) What are the image distances for objects placed at these distances from the lens: $5.00 \mathrm{~cm}, 8.00 \mathrm{~cm}$, $14.0 \mathrm{~cm}, 16.0 \mathrm{~cm}, 20.0 \mathrm{~cm} ?$ In each case, describe the image as real or virtual, upright or inverted, and enlarged or diminished in size. (b) If the object is $4.00 \mathrm{~cm}$ high, what is the height of the image for the object distances of $5.00 \mathrm{~cm}$ and $20.0 \mathrm{~cm} ?$

A diverging lens has a focal length of $-8.00 \mathrm{~cm}$.
(a) What are the image distances for objects placed at these distances from the lens: $5.00 \mathrm{~cm}, 8.00 \mathrm{~cm}$, $14.0 \mathrm{~cm}, 16.0 \mathrm{~cm}, 20.0 \mathrm{~cm} ?$ In each case, describe the
image as real or virtual, upright or inverted, and enlarged or diminished in size. (b) If the object is $4.00 \mathrm{~cm}$ high, what is the height of the image for the object distances of $5.00 \mathrm{~cm}$ and $20.0 \mathrm{~cm} ?$

Key Concepts

Thin Lens Equation

The thin lens equation (1/f = 1/d? + 1/d?) is a fundamental relation in geometrical optics that links the focal length of a lens with the distances of the object and the image from the lens. This formula is essential for calculating any one of these parameters when the other two are known, and it serves as the basis for understanding how lenses form images.

Focal Length Sign Convention

The sign of the focal length in lens problems indicates the type of lens. A positive focal length is associated with converging lenses (convex), while a negative focal length denotes diverging lenses (concave). This sign convention also helps determine whether the resulting image is real or virtual, and it is critical when applying the thin lens equation and magnification formulas correctly.

Diverging Lens Characteristics

Diverging lenses always produce images that are virtual, upright, and diminished relative to the object. Unlike converging lenses, which can produce inverted and real images under certain conditions, diverging lenses do not focus light to a real point; instead, the light rays appear to diverge from a point behind the lens, making the image accessible only when viewed through the lens.

Image Distance and Magnification

The image distance, determined from the thin lens equation, tells us the location of the image with respect to the lens, while the magnification (m = -d?/d?) provides information about the image's orientation and size relative to the object. A negative magnification indicates an inverted image, whereas a positive value indicates an upright image. For diverging lenses, the magnification is typically less than one in absolute value, indicating that the image is diminished.

Transcript

00:01 Here we're going to look at the imaging by a diverging lens.

00:06 A diverging lens has a negative focal length.

00:12 What you use for the imaging, there are two equations that you use for imaging by a lens.

00:20 One is called the thin lens equation.

00:25 And for that equation, we have that one over p, the object distance, plus one over q, the image distance is one over the focal length.

00:38 Do note that we have a negative focal length for a diverging lens.

00:47 And we will see that that results in a virtual or negative image distance of virtual image.

00:58 The other equation is magnification is the height of the image over the height of the object.

01:07 And that will be negative if the image is upside down.

01:12 It is also related to the q and the p as minus q over p.

01:21 Now we have five different p's, five centimeters, 14 centimeters, 16 centimeters, and 20 centimeters.

01:32 I have used a spreadsheet to calculate the image distance q.

01:38 So we have five of them, and i am showing a, b, not b, sorry, b is missing, and we'll do that one by hand.

01:54 We have a, c, d, and e in each of the rows.

02:01 And what we can see is that all the cues are negative, and that means they are all virtual.

02:14 And notice that the calculation of the magnification shows that they are all less than one and positive.

02:22 So this means that they are upright or non -inverted is a better way to say it.

02:30 And they are reduced in size...

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