A standard " 35-mm " slide measures 24.0 mm by 36.0 mm. Suppose a slide projector produces a 6

A standard " 35-mm " slide measures 24.0 mm by 36.0 mm....

A standard " $35-\mathrm{mm}$ " slide measures $24.0 \mathrm{~mm}$ by $36.0 \mathrm{~mm}$. Suppose a slide projector produces a $60.0-\mathrm{cm}$ by $90.0-\mathrm{cm}$ image of the slide on a screen. The focal length of the lens is $12.0 \mathrm{~cm}$. (a) What is the distance between the slide and the screen? (b) If the screen is moved farther from the projector, should the lens be moved closer to the slide or farther away?

Key Concepts

Thin Lens Equation

The thin lens equation relates the object distance, image distance, and focal length through the relationship 1/s + 1/s' = 1/f. This equation is fundamental in geometrical optics as it allows one to determine how far the lens should be from the object or the image to produce a clear, focused image, which is critical in projection systems and many optical instruments.

Magnification

Magnification is defined as the ratio of the image size to the object size and is also expressed as the negative ratio of the image distance to the object distance (m = -s'/s). This concept is essential for determining how much larger or smaller the projected image is compared to the actual slide or object, and understanding this relationship helps in designing and adjusting optical setups.

Adjustment of Lens Position Relative to Object and Screen

In optical projection systems, changing the distance between the image plane (screen) and the lens affects the required position of the lens relative to the object (slide) to maintain focus. This concept is crucial when the screen is moved farther from or closer to the projector, as it necessitates a corresponding adjustment in the lens position to compensate for the change in the image distance while still satisfying the thin lens equation.

Transcript

00:01 There are two equations that you use with thin lenses in order to figure out the imaging properties of that lens and what an image of a particular object looks like.

00:15 The first equation is called the thin lens equation.

00:23 And it basically relates the image distance from the lens called p, one over p, to the image distance, which is formed on a screen in this case, being a real image, but it relates the p and the q to one over the focal length of the lens.

00:47 But that's called the thin lens equation.

00:51 The second equation that is often useful, almost always, is an equation for the magnification.

01:00 It is given as the image height over the object height actual measurements there, or as it relates to the image and object distances, it is minus q over p.

01:20 So those two equations are used.

01:24 Now here we have a slide projector, which is basically a box with a movable converging lens, and you load your slides into the light source.

01:37 Up.

01:38 Down and the lens will create a real image on a screen some distance away.

01:47 But you have some control over the p and thus the q can be changed accordingly as well as we will see.

01:56 But the first thing that we will need to do is just work with the equations a little bit.

02:02 We're told that the focal length of the lens is 12 centimeters or 0 .12 .2.

02:09 We are told that it is a, let's see, it is a 36 by 24 millimeter slide.

02:23 Yeah, the slide projectors were called 35 millimeters, but a little bit of extra room there.

02:32 And the image, so this is the object, and the image is formed on the screen in the size of 90 by 60 centimeters.

02:48 Okay, so a reasonable size to view.

02:54 What we can use the information given is we can use it to determine the magnification.

03:04 So our equation number one, we can write that down as one over p plus one over q equals 1 over 0 .12.

03:20 What we can use our magnification information for is we can take our 90 centimeters, for example, and divide by the larger dimension, 3 .6 centimeters is the same thing as 36 millimeters.

03:41 And that is our magnification, and it turns out to be 25.

03:48 It will be the same for the other dimension, but you should probably check that.

03:54 And this is minus q over p.

03:59 So we basically have two equations, which we can use algebraically to figure out the q and the p.

04:10 So i would probably use the magnification equation to solve for say q in terms of p.

04:23 Q is minus 25p.

04:30 And then, okay, so the next step is to, use q equals minus 25 p in equation one...

Please give Ace some feedback