A converging lens (f1=24.0 cm) is located 56.0 cm to the left of a diverging lens (f2=-28.0 cm)

step 1 So we have two lenses. First we have lens L1 and for this the focal length is 24 centimeters, 24

A converging lens (f1=24.0 cm) is located 56.0 cm to the...

A converging lens $\left(f_{1}=24.0 \mathrm{cm}\right)$ is located $56.0 \mathrm{cm}$ to the left of a diverging lens $\left(f_{2}=-28.0 \mathrm{cm}\right) .$ An object is placed to the left of the converging lens, and the final image produced by the two-lens combination lies $20.7 \mathrm{cm}$ to the left of the diverging lens. How far is the object from the converging lens?

Key Concepts

Thin Lens Equation

The thin lens equation, expressed as 1/f = 1/d? + 1/d?, relates the focal length of a lens to the distances of the object (d?) and the image (d?) from the lens. This fundamental formula is used to determine one of these quantities when the other two are known, making it essential for analyzing image formation in optical systems.

Sequential Imaging Principle

In systems involving multiple lenses, the image produced by the first lens serves as the object for the second lens. Correctly and sequentially applying the thin lens equation to each lens, while accounting for the spatial separation between them, is key to analyzing and solving for the positions of final images or objects.

Sign Conventions in Optics

Proper application of sign conventions is critical in optics problems. These conventions designate positive or negative values for distances and focal lengths depending on whether the distances are measured in the direction of incoming light or opposite to it, and whether the lens is converging or diverging. This ensures that the mathematical treatment of the lens equation accurately reflects the physical configuration.

Optical System Geometry

Understanding the geometry of the optical system is important, especially when multiple lenses are involved. This involves accurately setting up coordinate systems, accounting for distances between lenses, and translating coordinates from one lens to another. Such geometric considerations allow for the correct formulation of equations that describe the relationships between object positions, image positions, and the spacing between optical elements.

Transcript

00:01 So we have two lenses.

00:03 First we have lens l1 and for this the focal length is 24 centimeters, 24 .0 centimeters.

00:12 Then we have another l2, lens 2, which is a diverging lens.

00:18 So the focal length for this is given as negative 28 .0 centimeters.

00:22 And the two lenses are a distance 56 .0 centimeters apart.

00:29 And an object is placed some distance which is unknown, x from lens 1, and the final image which is formed, that final image is 20 .7 centimeters from l2.

00:46 So this is what we do know.

00:49 So let's start talking about lens l1.

00:52 So for lens l1, what we do know is that the distance of the object is x and we do not know the distance of the image but we do know the focal length is 24 centimeters so let's start with the thin lens formula 1 over do plus 1 over di i is equal to 1 over f so 1 over x plus 1 over di is equal to 1 over 24 uh this is a addition of two fractions so this is a equals x plus di or di i plus x over x times di is equal to 1 over 24 and then we can cross multiply so that could give you 24 times di plus 24 times x is equal to x times di so now we can gather terms which have di in them together, and that would give me 24 di minus x, di is equal to negative 24x.

02:09 So if you factor the i, that would give you 24 minus x is equal to negative 24x, or you can say, di is equal to negative 24x over 24 minus x.

02:27 To remove the negative sign, i'm just going to switch the positions of the two terms in the denominator.

02:35 So the i'm going to say is equal to 24x over x minus 24...

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