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

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

Sign Conventions

Correct use of sign conventions is crucial in lens problems. For lenses, distances measured in the direction of the incoming light are typically considered positive, while distances measured in the opposite direction are negative. The focal length of a diverging lens is taken as negative, and this sign informs the nature of the image (virtual vs real) formed by the system.

Magnification

Magnification in lens systems is defined as the ratio of the image height to the object height (or equivalently, the negative ratio of the image distance to the object distance when using the thin lens formula). The sign of the magnification indicates whether the image is inverted (negative value) or upright (positive value), while its magnitude indicates whether the image is enlarged or diminished relative to the object.

Thin Lens Equation

The thin lens equation, given by 1/f = 1/do + 1/di, is fundamental for analyzing image formation by lenses. It relates the focal length (f) of the lens, the object distance (do), and the image distance (di), allowing one to determine one quantity if the other two are known. This equation is applicable to both converging and diverging lenses, provided the proper sign conventions are used.

Diverging Lenses

Diverging lenses are characterized by their negative focal length, meaning that they cause light rays to spread apart rather than converge. They always form virtual, upright, and diminished images regardless of the object position. Understanding the behavior of diverging lenses is essential when working with problems involving image formation where the lens does not bring rays to a real focus.

Transcript

00:01 For this problem, we're considering a diverging lens with focal length of negative 8 centimeters.

00:10 So that means that f is going to be equal to negative 8 centimeters.

00:22 And for this lens, we want to find the image distance for a variety of different object distances.

00:33 And determine whether the image will be real or virtual, upright or inverted, enlarged or diminished.

00:42 We then want to consider for part b if there's an object with a height of four centimeters, what the height of the image would be for a distance of five centimeters and 20 centimeters.

01:00 So in order to do this, we want to start with the equation that 1 over p plus 1 over q equals 1 over f, where p is the object distance, q is the image distance, and f is the focal length.

01:17 For part a, we're going to be solving for q, so we're going to do a little bit of algebra to rearrange that equation, and we would get that q.

01:31 Is equal to minus 1 eighth, because i've plugged in negative 8 centimeters for the focal length, minus 1 over p to the negative first power, so to the inverse.

01:49 And that is the equation we're going to be using for each of the distances, p, in part a.

01:59 So for 5 centimeters, when you plug in 5 centimeters, you would get that q is negative 3 centimeters.

02:08 When you plug in 8 centimeters for p, you get that q is negative 4 centimeters.

02:16 And i'm rounding to the nearest centimeter here.

02:21 You may not want to round in your answers.

02:24 But continuing on, for p equal to 14 centimeters, q will be negative 5 centimeters.

02:33 For 16, q is also about negative 5 centimeters, and when the object is at a distance of 20 centimeters way, the image will be at a distance of negative 6 centimeters.

02:54 Because this is a diverging lens, all of these images will be virtual.

03:02 None of them will be real...

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