An object is located 30.0 cm to the left of a converging lens whose focal length is 50.0 cm . (a

An object is located 30.0 cm to the left of a converging...

An object is located $30.0 \mathrm{cm}$ to the left of a converging lens whose focal length is $50.0 \mathrm{cm} .$ (a) Draw a ray diagram to scale and from it determine the image distance and the magnification. (b) Use the thin-lens and magnification equations to verify your answers to part (a).

An object is located $30.0 \mathrm{cm}$ to the left of a converging lens whose focal length is $50.0 \mathrm{cm} .$ (a) Draw a ray diagram to scale and from it determine the image distance and the magnification.
(b) Use the thin-lens and magnification equations to verify your answers to part (a).

Key Concepts

Converging Lenses

Converging lenses, also known as convex lenses, are optical devices that bend parallel incoming light rays towards a single point called the focal point. They are fundamental in geometric optics for forming images of objects by converging light, and their behavior is characterized by their focal length, which determines how strongly they converge the light rays.

Ray Diagrams in Geometric Optics

Ray diagrams are graphical tools used in geometric optics to determine the path of light rays through lenses or mirrors and the resultant image properties. By drawing principal rays—such as rays parallel to the optical axis, rays through the focal point, and rays through the center of the lens—one can visually locate the position and size of the image and appreciate how the lens forms the image.

Thin-Lens Equation

The thin-lens equation, given by 1/f = 1/do + 1/di, is a fundamental relationship in optics that links the focal length of a lens (f) with the object distance (do) and the image distance (di). This equation is essential for quantitatively analyzing the imaging properties of lenses and predicting where the image will form relative to the lens.

Magnification in Optics

Magnification is a measure of how much larger or smaller an image is compared to the object, usually expressed as m = -di/do. The negative sign indicates image inversion for real images formed by converging lenses. Magnification reveals not only the size relationship between the image and the object but also provides insight into the image orientation relative to the object.

Transcript

00:01 Yeah, since the focal length and distance mentioned in the problem or large, we are going to start by selecting a scale so a good appropriate scale would be escape wants to fight, which means that one centimeter on the diagram represents five centimeters in real.

00:19 So we start by drawing a horizontal line, and on that horizontal line we draw vertical line.

00:25 This vertical line will represent the lens report that the lens has a focal length off 50 centimeters, so on a scale of one is to fight.

00:35 That would be 10 centimeters.

00:37 So we measure off 10 centimeters from the lands and put the talk there.

00:42 This doctor presents the focal length, then you distance off six centimeters, which represents 30 centimeters in actual, and we please our object there because the question says the object is 30 centimeters from the lens.

00:58 Now that dr diagrams all set up, we will horizontal line from the top off the object to the lens.

01:06 This is a real flight, and then after the rio flight passes through the lens, it will go on to pass through the focal point on the other side.

01:15 So that's how the line will continue.

01:17 Then we draw another line again, starting from the top off.

01:20 The object will go through the center of the lens and this rate will not be deflected.

01:28 So we realized that the to raise our now divergent they will not intersect...

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