Two systems are formed from a converging lens and a...

Two systems are formed from a converging lens and a diverging lens, as shown in parts $a$ and $b$ of the drawing. (The point labeled $" F_{\text {converging }} "$ is the focal point of the converging lens.) An object is placed inside the focal point of lens 1 at a distance of $10.00 \mathrm{~cm}$ to the left of lens 1 . The focal lengths of the converging and diverging lenses are $15.00$ and $-20.0 \mathrm{~cm}$, respectively. The distance between the lenses is $50.0 \mathrm{~cm}$. Determine the final image distance for each system, measured with respect to lens 2 .

Key Concepts

Lens Equation

The lens equation, given by 1/f = 1/d? + 1/d?, is a fundamental relation in geometric optics that links an object’s distance (d?), the image’s distance (d?), and the lens's focal length (f). It is the primary tool for determining where an image will be formed by a lens, and whether the image is real or virtual, based on the object distance relative to the lens's focal point.

Sign Conventions in Optics

Sign conventions are crucial in lens calculations, as they assign positive or negative values to distances and focal lengths depending on their orientation relative to the lens. This systematic approach helps in correctly identifying the nature of the image—real images typically have positive image distances while virtual images have negative image distances, with the focal length being defined as positive for converging lenses and negative for diverging lenses.

Image Formation by Lenses

Image formation involves the bending of light rays as they pass through a lens, causing convergence or divergence that results in an image. The characteristics of the formed image, such as its size, orientation, and whether it is real or virtual, depend on both the type of lens (converging or diverging) and the distance of the object from the lens relative to its focal point.

Multiple-Lens Systems

When dealing with systems containing more than one lens, the image produced by the first lens acts as the object for the subsequent lens. This sequential process requires careful application of the lens equation for each lens, taking into account the distance between the lenses. This concept is pivotal for understanding complex optical setups used in devices like telescopes and microscopes.

Transcript

00:01 Two systems are formed from a converging lens and a diverging lens, as shown in parts a and b of the figure in your textbook.

00:10 The point labeled f converging is the focal point of the converging lens.

00:14 An object is placed inside the focal point of lens 1 at a distance of 10 centimeters, so the left of lens 1.

00:20 Focal lengths of the converging and diverging lenses are 15 and negative 20 centimeters respectfully.

00:27 The distance between the lenses is 50 centimeters.

00:29 Determine the final image distance for each system measured with respect to lens 2.

00:36 Okay, so start off with our thin lens equation.

00:40 1 over f is equal to 1 over do plus 1 over d .i.

00:46 We have lots of solving for dis here, so let's just go ahead and rewrite this as 1 over f minus 1 over d .o, all of us to the negative 1 power.

00:58 So for part a, you have a focal length of 15 centimeters, and you have do is 10 centimeters, plugging this in, and this gives us a d .i of negative 30 centimeters.

01:22 So the distance of the object from the diverging lens, do is negative d .o.

01:30 Is negative d .i.

01:31 Plus d.

01:32 So negative negative 30 is positive 30 plus the distance, which is 50 as given in the problem.

01:39 So this is 80 centimeters.

01:45 So doing this again for the other lens...

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