Problem 80 - College Physics

Question

Two lenses, each having a power of \(+10 \mathrm{D}\), are placed \(20 \mathrm{~cm}\) apart along the same axis. If an object is \(60 \mathrm{~cm}\) from the first lens (not in between the lenses), where is the final image relative to the first lens, and what are its characteristics?

Step-by-Step Solution

Answer

The final image is located 24.44 cm from the first lens; it is real and inverted.

Step 1: Understand the Problem

Two lenses with a power of \(+10 \mathrm{D}\) each are placed \(20 \mathrm{cm}\) apart. The object is located \(60 \mathrm{cm}\) from the first lens. We need to find the position of the final image relative to the first lens and its characteristics.

Step 2: Calculate the Focal Length

The power of each lens is given as \(+10 \mathrm{D}\). The formula to convert power to focal length is \[ f = \frac{1}{P} \] Thus, the focal length \( f \) for each lens is \[ f = \frac{1}{10} = 0.1 \text{ meters} = 10 \text{ cm} \]

Step 3: Use Lens Formula for First Lens

The lens formula is \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] For the first lens, \(f = 10 \text{ cm}\) and the object distance \(u = -60 \text{ cm}\) (object distance is taken negative based on sign convention). We need to find the image distance \(v\).\[ \frac{1}{10} = \frac{1}{v} + \frac{1}{60} \]Solving for \(v\):\[ \frac{1}{v} = \frac{1}{10} - \frac{1}{60} = \frac{6 - 1}{60} = \frac{5}{60} \]\[ v = \frac{60}{5} = 12 \text{ cm} \]The image formed by the first lens is \(12 \text{ cm}\) on the positive side (real image), \(12\text{ cm}\) from the lens.

Step 4: Calculate Object Distance for Second Lens

The distance between the two lenses is \(20 \text{ cm}\), so the object distance for the second lens \(u'\) is \[ u' = 20 - 12 = 8 \text{ cm} \]This is because the image formed by the first lens becomes the object for the second lens.

Step 5: Use Lens Formula for Second Lens

The second lens also has \(f = 10 \text{ cm}\). Now, with the object distance \(u' = 8 \text{ cm}\), the lens formula is applied:\[ \frac{1}{10} = \frac{1}{v'} - \frac{1}{8} \]Solving for \(v'\):\[ \frac{1}{v'} = \frac{1}{10} + \frac{1}{8} = \frac{4 + 5}{40} = \frac{9}{40} \]\[ v' = \frac{40}{9} \approx 4.44 \text{ cm} \]The image is formed \(4.44 \text{ cm}\) on the positive side from the second lens.

Step 6: Calculate Final Image Position Relative to First Lens

To find the final image position relative to the first lens, add \(20 \text{ cm}\) (distance between lenses) to \(4.44 \text{ cm}\) (distance from the second lens):\[ 20 + 4.44 = 24.44 \text{ cm} \]Therefore, the final image is located \(24.44 \text{ cm}\) from the first lens, on the same axis as the lenses.

Step 7: Determine Image Characteristics

The image is formed on the positive side for both lenses, indicating it is a real and inverted image relative to the object, and it is located \(24.44 \text{ cm}\) from the first lens.

Key Concepts

Lens Power

Lens power is a measure of how much a lens can bend light-rays. It's expressed in diopters (D) and calculated as the reciprocal of the focal length in meters. The formula for power is given by:

• Power (\( P \)) = \( \frac{1}{f} \)

Here, \( f \)is the focal length measured in meters.
This means that a lens with a power of \(+10 \text{ D} \)can bend light strongly, focusing parallel rays of light to a point 10 centimeters away from the lens.
High power lenses are typically used in eyeglasses and optical devices that require a strong focus, such as microscopes and binoculars.
Lenses with positive power converge light to form real images, whereas negative power lenses diverge light and usually form virtual images.

Focal Length Calculation

Understanding how to calculate the focal length is key to studying optics. When we talk about a lens, the focal length is the distance from the lens to the focal point where light rays converge.
The focal length is crucial because it indicates the strength of the lens.
The calculation of the focal length from lens power is straightforward:

• For any given lens power, use the formula: \[ f = \frac{1}{P} \]

For example, if the lens has a power of \(+10 \text{ D} \), like in our original problem, the focal length would be \(10\text{ cm} \).
This helps in determining how lenses can be arranged in optical instruments to form clear images.
Keep in mind that the focal length is positive for converging lenses and negative for diverging ones.

Lens Formula

The lens formula plays a fundamental role in determining the image distance formed by a lens. It relates the object distance, image distance, and the focal length. The formula is:

• \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)

Where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance.
This formula helps us find out where an image will be located after passing through a lens.
Its application is crucial in designing camera lenses, eyeglasses, and other optical instruments where precise imaging is required.
Note that distances measured in the direction of incident light are considered positive, while those against it are negative.

Real Image Characteristics

Real images are formed when light rays converge and actually cross at the image point, different from virtual images where rays only appear to converge.
For a real image:

• It’s usually formed on the opposite side of the source light in a lens setup.
• It can be projected onto a screen.
• It's inverted compared to the object.

In our original exercise, both lenses form real images.
The noteworthy trait of real images is their ability to be captured on a surface, which is essential in photography and cinematography.
Objects seen through devices like cameras or telescopes typically produce real images, allowing detailed observation and evaluation. Understanding these characteristics helps in various practical applications, from creating focused pictures to complex scientific observations.