Show that identical objects placed equal distances on either...

Key Concepts

Mirror and Lens Equations

These equations, typically written as 1/f = 1/do + 1/di, relate the focal length of an optical instrument to the distances of the object and image from the device. They are fundamental in predicting how an object placed at a certain distance from a mirror or lens will produce an image with specific properties, such as size and position.

Magnification

Magnification describes the ratio of the image size to the object size and is directly linked to the object and image distances. In optical systems, when objects are placed symmetrically about a key point like the focal point, the magnification magnitudes can become identical, leading to images of equal size despite potential differences in orientation or other qualities.

Real and Virtual Images

Optical systems such as concave mirrors and converging lenses can produce either real images—formed when light converges and can be projected—or virtual images—formed when light appears to diverge from a point behind the mirror or lens. The nature of the image (real or virtual) depends on the object's placement relative to the focal point, so while image sizes may be equal due to symmetrical placement, the images themselves can be of opposite types.

Transcript

00:01 Problem 73.

00:02 Show that identical objects placed equal distances on either side of the focal point of the concave mirror or converging lens produce images of equal size.

00:11 And then they ask are the images of the same type.

00:13 So we need to find the images or at least create an expression for the images when they're at position 1, which is x units away from the focal length to the left.

00:22 And position 2, which is x units away from the focal length on the right.

00:26 So we'll start with position 1, where my object distance, s, is the focal length plus x.

00:35 I'm going to use the mirror equation, 1 over f equals 1 over s plus 1 over s prime, and basically substitute in what my s is, my s is f plus x.

00:53 And then i can solve this equation, just do one more arrow to the right here.

00:58 One over s prime is going to be one over f minus one over plus x.

01:06 So i can just do a little algebra and get this so that it's solved for s prime.

01:13 I need a common denominator.

01:15 So i'll keep it in this form for now.

01:18 If i make the first term equal to one over for f, i just multiply top and bottom by f plus x.

01:28 And if i make the second term, whoops, i don't want a positive there.

01:32 I want a negative.

01:35 I can just multiply top and bottom by f so that i have a common denominator.

01:42 F over f times f plus x is the same thing as one over f plus x.

01:45 So now i have the same denominator for both terms.

01:49 So f plus x minus f, the fs are going to cancel, all divided by f times f.

01:57 Plus x and that's equal to 1 over s prime.

02:02 So you can see in the numerator my f goes away.

02:05 What i'm left with is 1 over s prime equals x over f f plus x and then therefore s prime is equal to f times f plus x over x.

02:22 So that's where i am at this point.

02:25 I'm going to do the same thing for position two and then we'll look at the magnifications because to find to show that they're the same size we can show that they have the same magnification so i'm going to do position two if you look at position two the the image distance is the focal length minus x so i'm going to do the same exact process but my s is the focal length minus x so going through that whole scenario again i have one over the focal length equals 1 over f minus x plus 1 over s prime.

03:03 I'm going to again solve this for s prime.

03:05 We can probably do it a little more quickly this time.

03:09 1 over f minus 1 over f minus x equals 1 over s prime.

03:16 Now my 1 over s prime is on the right.

03:19 I kind of want it on the left...

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