A 1.00-cm -high object is placed 4.00 cm to the left of a converging lens of focal length 8.00

A 1.00-cm -high object is placed 4.00 cm to the left of a...

Key Concepts

Thin Lens Equation

This equation relates the focal length of a lens with the distances of the object and image from the lens via the formula 1/f = 1/d_o + 1/d_i. It is a fundamental tool for calculating where an image will be formed when an object is placed at a known distance from a lens, and it applies to both converging and diverging lenses with appropriate sign conventions.

Magnification

Magnification describes how the size of an image relates to the size of the object and is given by the ratio m = h_i/h_o = -d_i/d_o. It not only indicates how much larger or smaller the image is compared to the object but also provides information about the image's orientation (inverted or upright) based on its sign.

Sign Conventions in Optics

Sign conventions are crucial when applying the lens equation and magnification formulas. They define which distances and focal lengths are considered positive or negative based on the lens type and the directions of the incoming and outgoing light rays. For instance, converging lenses have positive focal lengths while diverging lenses have negative ones.

Compound Lens Systems

In systems with multiple lenses, the image formed by one lens becomes the object for the next. Analyzing compound lens systems requires sequential application of the thin lens equation and the magnification formula for each lens, taking careful note of distances and sign conventions between the lenses.

Real vs Virtual Images

Real images are those formed by actual convergence of light rays and can be projected onto screens, while virtual images arise when the light rays only appear to diverge from a point and cannot be projected. Determining whether an image is real or virtual involves analyzing the geometry and sign of the image distance.

Image Orientation

The orientation of an image—whether it is inverted or upright—is determined by the sign of the magnification. A negative magnification indicates an inverted image, while a positive magnification indicates an upright image. This concept is essential when interpreting the nature of images formed by various optical instruments.

Transcript

0:00 High there.

00:01 So for this problem, we have a one centimeter high object.

00:05 So we are given the object distance b1.

00:10 That is placed.

00:17 Okay, so the value given is four centimeters.

00:21 And it is one centimeter high.

00:27 And to the left of a converging lens with a focal length that is also given and that is equal to a centimeter.

00:37 So a diverging length of focal length, 16 centimeters.

00:50 We're given that the focal distance 2 that we're going to call 2 is minus 16 centimeters and is 6 centimeters to the right of the converging lens.

01:04 So we will have that p2, if i'm correct.

01:11 Is going to be the sum of those six centimeters plus the value that we obtain for the image distance q1.

01:26 So with that said, what we need to determine is to find the position in height of the final image, and to answer the question, is the image inverted or upright, real or virtual? so to answer this question, we first start with the thin lens equation that states that the object, the object image distance is equal to the product between the object position or the object distance times the focal distance 1 divided by the difference between these two values.

02:09 So to obtain this, we just simply substitute those values.

02:12 We have four centimeters.

02:13 For p1 and the focal distance 1 is 8 centimeters and this divided by the difference between these two.

02:21 So we will have 4 centimeters minus 8 centimeters.

02:25 So from this we obtain minus 8 centimeters.

02:34 So that is the object distance 1 and the image distance 1.

02:40 And the magnification by the first length is going to be minus q1 over p1.

02:49 So we put those values in here.

02:51 We will have that this is minus minus 8 centimeters that we obtain divided by 4 centimeters.

02:59 So from this we obtain a magnification, a positive modification of 2.

03:04 So this virtual image formed by the first lens is the object for the size.

03:10 Length.

03:12 So we're going to have that.

03:14 P2, as i said, is the sum of the six centimeters given plus q1...

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