A converging lens with a focal length of 3.00 cm is placed 24.00 cm to the right of a concave

A converging lens with a focal length of 3.00 cm is placed...

A converging lens with a focal length of $3.00 \mathrm{~cm}$ is placed $24.00 \mathrm{~cm}$ to the right of a concave mirror with a focal length of $4.00 \mathrm{~cm}$. An object is placed between the mirror and the lens, $6.00 \mathrm{~cm}$ to the right of the mirror and $18.00 \mathrm{~cm}$ to the left of the lens. Name three places where you could find an image of this object. For each image tell whether it is inverted or upright and give the total magnification.

Key Concepts

Lens Equation

The lens equation, represented as 1/f = 1/d_o + 1/d_i, is used to determine the position and characteristics of an image formed by a thin lens. It relates the focal length of the lens to the distances of the object and the image from the lens, providing a mathematical description of how converging or diverging lenses form images.

Mirror Equation

The mirror equation, similar to the lens equation, is given by 1/f = 1/d_o + 1/d_i and is used to analyze image formation in spherical mirrors. It connects the focal length of the mirror with the object and image distances, allowing one to determine whether the image will be real or virtual and its relative position to the mirror.

Magnification

Magnification quantifies the size and orientation of the image relative to the object. It is calculated as the negative ratio of the image distance to the object distance, where a negative magnification indicates an inverted image and a positive value denotes an upright image. This concept is crucial for understanding how optical systems alter the apparent size of objects.

Composite Optical Systems

In systems involving more than one optical element, such as a lens and a mirror, the image formed by one component can act as the object for the next. Analyzing such composite systems requires sequential application of the lens and mirror equations along with careful tracking of distances and orientation to accurately predict where multiple images will appear and their characteristics.

Transcript

00:03 In this problem we have a concave mirror and a converging lens that are 24 series apart.

00:12 We also have an object that is placed six ceramic mirrors in front of the mirror.

00:19 And we want to compute three different places where this object forms an image.

00:27 Let's start with the image produced by the mirror.

00:31 So for the mirror, we can use the mirror equation 1 over f that is the focal length but for the mirror this is f1 and we know from the information of the problem that is f1 is 4 30 meters okay so the equation is this and we are calling p1 and q1 the object and the image distances just because of the mirror so if we replace the values here we have 1 over 4 equal 1 over p1 and p1 is 6 thermometers.

01:17 So it is, this is the distance, the object distance for the mirror.

01:23 And from this, we can compute the value of q1.

01:30 So q1 is just 12 .0 centimeters.

01:44 We can also compute the magnification.

01:47 So the magnification is minus q1 over p1 and replacing the values, this is just 12 and p1 is 6.

01:59 So this is minus 2 .00.

02:05 Okay, so the first image is produced just by the mirror and is located at 12 centimeters in front of the mirror.

02:15 Positive, the image is positive and that means that is located in front of the mirror.

02:24 So it's more or less in this point and all of this is 12 meters.

02:34 And as the magnification is negative, this is an invert image.

02:40 So it's maybe something like this.

02:47 The magnification, the absolute value is too, so that size of the image is twice the size of the object.

02:57 Okay, so this is the first image.

03:00 The second image is the image that is produced by the lens, only by the lens.

03:07 And for the lens, we can use also the lens equation that is one over f2, f2 because f2 is the focal length of the lens is equal 1 over p2 plus 1 over q2.

03:25 So replacing the values here for f2 is 3.

03:31 For p2, p2 is just the object distance.

03:35 The original object is at this point here.

03:41 So the object distance for the length is just 24 centimeters minus 6 cemeter.

03:49 All of this is 18.

03:51 18 cemeter is here.

03:55 So it's just 1 over 8...

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