A slide projector uses a lens of focal length 115 mm to focus a 35 mm slide (having dimensions 2

step 1 Okay guys, let's take a look at the first question. So the first question was asking us, where s

A slide projector uses a lens of focal length 115 mm to...

A slide projector uses a lens of focal length 115 $\mathrm{mm}$ to focus a 35 $\mathrm{mm}$ slide (having dimensions 24 $\mathrm{mm} \times 36 \mathrm{mm} )$ on a screen. The slide is placed 12.0 $\mathrm{cm}$ in front of the lens. (a) Where should you place the screen to view the image of this slide? (b) What are the dimensions of the slide's image on the screen?

Key Concepts

Thin Lens Equation

The thin lens equation, given by 1/f = 1/object distance + 1/image distance, is fundamental in geometrical optics for relating the focal length of a lens to the distances of the object and image from the lens. This relation allows one to determine where an image will be formed when an object is placed at a known distance from a converging or diverging lens.

Magnification

Magnification in optics is defined as the ratio of the image size to the object size and can also be expressed as the ratio of the image distance to the object distance. This concept not only tells us by what factor the image has been enlarged or reduced but also indicates that the image produced by a converging lens is inverted relative to the object when the image distance is positive.

Real Image Formation

Real image formation occurs when light rays converge after passing through a lens, producing an image that can be projected onto a screen. In the context of optical devices like projectors, understanding real image formation is essential to correctly position the screen and determine the size of the image, ensuring the proper focus and clarity of the image output.

Transcript

00:01 Okay guys, let's take a look at the first question.

00:02 So the first question was asking us, where should you place the screen to view the image of the slide? which was asking us, what is the image distance for this question? okay? so we know that 1 over focal lens is equal to 1 over image distance minus 1 over object lens.

00:25 F is the focal length, which is given as 115 millimeter.

00:29 U is the object distance, which is given as 12 .0 centimeter.

00:33 So now we know that v is equal to uf over u plus f.

00:37 If we plug in the value we'll get 12 .0 centimeter is 120 millimeter plus i'm sorry times 115 millimeter divided by 120 millimeter plus 115 millimeter and this will give us the image distance is about 58 .7 millimeter which is 5 .87 centimeter okay so if we draw a diagram if we say this is the the length okay and if we say that uh you were saying that the the slides place 12 centimeter in front of the lens okay so we say that the slice is right here okay, this is the slice and distance between it are 12 .0 centimeter.

02:06 So now since we got the image distance is given as 58 .7 millimeter, which is 5 .87 millimeter.

02:17 That means we need to place the screen somewhere like here.

02:24 Let's say the straight line here is the screen, this is the screen.

02:29 And this is 5 .87 centimeter.

02:36 Okay, so we should place the screen behind the lens and which is about 5 .8, sorry, 5 .87 centimeter away from the lens.

02:53 So for question b, it was asking us what is the dimension of the slides image on the screen? while we know this is asking us to find out the height of the size image on the screen, image...

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