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

step 1 This is a problem involving a diverging lens. You're asked to find the image distance, Q, given

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

Key Concepts

Diverging Lens

A diverging lens is a type of lens that has at least one surface that curves inward, causing incident parallel light rays to spread apart (diverge) after refraction. Its defining characteristic is that it has a negative focal length, meaning that light appears to emanate from a focal point on the same side of the lens as the incoming light. This optical behavior typically results in virtual, upright, and reduced images regardless of the object's position relative to the lens.

Lens Formula

The lens formula, expressed as 1/f = 1/do + 1/di, is a fundamental relationship in optics that connects the focal length (f) of the lens, the object distance (do), and the image distance (di). This equation is used to calculate any one of these parameters when the other two are known. For diverging lenses, the convention keeps the focal length negative, which influences the calculated image distance as well.

Sign Conventions

Sign conventions in optics are essential for properly applying the lens formula and interpreting the resultant image characteristics. For lenses, the object distance (do) is taken as positive if the object is placed on the same side as the incoming light, while the image distance (di) is positive for real images formed on the opposite side of the lens and negative for virtual images formed on the same side as the object. The focal length is positive for converging lenses and negative for diverging lenses.

Magnification

Magnification in lens systems is defined by the ratio m = -di/do, where m indicates how much larger or smaller the image is compared to the object. The negative sign reflects the inversion of the image for real images, while for virtual images produced by diverging lenses, the magnification will be positive, indicating that the image is upright relative to the object. This concept helps in determining the size of the image as well as its orientation.

Image Characteristics

The characteristics of an image formed by a lens describe whether the image is real or virtual, upright or inverted, and magnified or reduced. In diverging lenses, the images are typically virtual (meaning they cannot be projected on a screen because they require an observer to be seen), upright relative to the object, and diminished in size. Understanding these characteristics is important for practical applications in optics and imaging systems.

Transcript

00:01 This is a problem involving a diverging lens.

00:04 You're asked to find the image distance, q, given certain object distances, p in the problem.

00:15 You get five different distances.

00:17 I'll go with at least one solution.

00:18 Then you can use the same method for solving all five.

00:21 So you're going to want to solve for a q in the end.

00:23 So there's several different ways you can do this, plugging the number first and then the algebra.

00:27 I'm going to go ahead and do the algebra first on the equation.

00:31 And then use that to get the solution for at least one.

00:35 So i'm going to solve for q, the lens equation.

00:40 So first i subtract the one over p to the other side.

00:43 Common denominator on the left hand side would be fp.

00:46 And so you're going to have p over fp minus f over fp equals 1 over q, or p minus f over fp equals one over a cube and then you inverse both sides and you get f p over p minus f equals q so i'm going to use that equation you're given f in the problem and you're given as i said the various p values so i'm going to go ahead and plug in for the first one here f negative eight times five p over p5 minus minus 8 is going to equal q and so you get negative 40 over this becomes plus positive over 13 equals q or when you do the division there you get q is equal to 3 .08 centimeters equals q and so that would be the answer for the five centimeter distance now use the same approach of all the other answers.

02:08 And when i did that, i got the following numerical answers.

02:13 By the way, this should be negative, because the negative sign is still there.

02:17 So negative 3 .08.

02:19 So for the second one, when i solved it, i got negative 4 .00 centimeters.

02:25 For the 14 centimeter object distance, i got negative 5 .09 centimeters.

02:32 For the 16, i got negative 5 .09 centimeters.

02:36 Three centimeters and for the 20 i got negative 5 .71 centimeters and so these are the answers for each one of those object distances if you use the same approach as i did for the first one it also asks in that problem uh about the um it says in each case describe the images as real or virtual upright inverted and so on um because it's a diverging lens there's only one set of possibilities.

03:15 Diverging lens will always produce images that are virtual, upright, and diminished in size.

03:33 Very diminished...

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