Problem 92 - Physics for Scientists and Engineers with Modern Physics

Question

Two converging lenses are placed \(30.0 \mathrm{~cm}\) apart. The focal length of the lens on the right is \(20.0 \mathrm{~cm},\) and the focal length of the lens on the left is \(15.0 \mathrm{~cm} .\) An object is placed to the left of the \(15.0-\mathrm{cm}\) -focal-length lens. A final image from both lenses is inverted and located halfway between the two lenses. How far to the left of the \(15.0-\mathrm{cm}\) -focal length lens is the original object?

Step-by-Step Solution

Answer

The object is 20 cm to the left of the 15 cm-focal-length lens.

Step 1: Identify Known Variables

From the problem, we know:- The distance between the lenses: \(d = 30.0\,\mathrm{cm}\)- Focal length of the left lens: \(f_1 = 15.0\,\mathrm{cm}\)- Focal length of the right lens: \(f_2 = 20.0\,\mathrm{cm}\)- Final image is inverted and halfway between lenses, so at \(15.0\,\mathrm{cm}\) from each lens.

Step 2: Apply Lens Formula for the Right Lens

The lens formula is given by \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\).Since the final image is located halfway between the lenses, it is 15.0 cm to the left of the right lens. Therefore, \(v_2 = -15.0\,\mathrm{cm}\) (image distance is negative for virtual images) and \(f_2 = 20.0\,\mathrm{cm}\).Plug these into the lens formula—\(\frac{1}{20} = \frac{1}{(-15)} - \frac{1}{u_2}\)—to find \(u_2\).

Step 3: Calculate Image Distance for the Right Lens

From the formula \(\frac{1}{20} = \frac{1}{(-15)} - \frac{1}{u_2}\),\[\frac{1}{u_2} = \frac{1}{(-15)} - \frac{1}{20} = -\frac{4}{60} + \frac{3}{60} = -\frac{1}{60}\]Thus, \(u_2 = -60\,\mathrm{cm}\).

Step 4: Determine Object Distance for the Left Lens

Now, since \(u_2\) is the image distance for the left lens, this is also the value for \(v_1\), the image distance from the left lens, but changed in sign.Thus, \(v_1 = +60\,\mathrm{cm}\).Apply the lens formula for the left lens: \(\frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u_1}\).So, \(\frac{1}{15} = \frac{1}{60} - \frac{1}{u_1}\).

Step 5: Solve for Object Distance

Using \(\frac{1}{15} = \frac{1}{60} - \frac{1}{u_1}\),\[ \frac{1}{u_1} = \frac{1}{60} - \frac{1}{15} = \frac{1 - 4}{60} = -\frac{3}{60} = -\frac{1}{20} \]Thus, \(u_1 = -20\,\mathrm{cm}\). This means the object is placed 20 cm to the left of the left lens.

Key Concepts

Converging lenses

Converging lenses, often known as convex lenses, are designed to focus light rays passing through them. They have a distinct shape where the center is thicker than the edges, causing incoming parallel light rays to converge at a single point called the focal point. These lenses are used in various applications, from eyeglasses to camera lenses.
In optics, it's essential to understand that the focal length is the distance between the lens's center and its focal point. It determines how powerful the lens is in converging light rays. A shorter focal length means the lens is stronger, as it bends light more effectively.

• Converging lenses form images by refracting light twice, once upon entry and once upon exiting.
• The position and size of the image created depend on the object's distance from the lens and the focal length.
• Converging lenses can produce real, inverted images if the object is placed beyond the focal point.

Experimenting with converging lenses can help students better understand focal points and the concept of refraction.

Lens formula

The lens formula is a mathematical equation that relates three key components: the focal length of the lens, the object distance, and the image distance. This formula is crucial for solving many problems relating to lens optics and can be written as:\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]Here, \( f \) represents the focal length, \( v \) is the image distance from the lens, and \( u \) is the object distance from the lens. This formula helps predict where an image will form when an object is placed at a certain distance from a lens.
By rearranging the lens formula, it's possible to solve for any unknown variable if the other two are known. Understanding and applying this formula is crucial to solving most optics problems that involve lenses.

• The sign convention is important: for converging lenses, focal length \( f \) is positive.
• If the image forms on the opposite side of the object, \( v \) is positive, showing that the image is real.
• If the image is on the same side as the object, \( v \) is negative, indicating a virtual image.

Mastery of the lens formula allows students to understand how different variables interact in lens systems.

Image distance

Image distance refers to the distance from the lens to the image that the lens forms. Together with the object distance and the focal length, it helps predict the characteristics of the image, such as its location, size, and orientation.
In exercises involving converging lenses, like the one we are examining, determining the image distance is a vital step, as it faciliates understanding of where the image forms in relation to the lens.

With each lens adjustment, particularly in systems with multiple lenses:

• The image from the first lens becomes the object for the second lens.
• The position and nature (real or virtual) of this "intermediate" image affect the final image.
• Using the lens formula helps to find the image distance; a positive distance indicates a real image, while a negative one indicates a virtual image.

In the given problem, using the image distance helps to understand how each lens contributes to the final image, showcasing the importance of handling each lens calculation with precision.