Problem 41 - University Physics with Modern Physics

Question

Combination of Lenses II. Two thin lenses with a focal length of magnitude \(12.0 \mathrm{cm},\) the first diverging and the second converging, are located 9.00 \(\mathrm{cm}\) apart. An object 2.50 \(\mathrm{mm}\) tall is placed 20.0 \(\mathrm{cm}\) to the left of the first (diverging) lens. (a) How far from this first lens is the final image formed? (b) Is the final image real or virtual? (c) What is the height of the final image? Is it erect or inverted? (Hint: See the preceding two problems.)

Step-by-Step Solution

Answer

The final image is 17.74 cm to the right of the first lens, real, 0.84 mm tall, and erect.

Step 1: Determine the Image Distance from the Diverging Lens

The formula for the lens is given by \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). The focal length \( f \) for the first lens is \(-12.0 \mathrm{cm}\) (diverging). The object distance \( d_o \) is \(20.0 \mathrm{cm}\). We rearrange the formula to solve for the image distance \( d_i \) from the first lens:\[ \frac{1}{-12.0} = \frac{1}{20.0} + \frac{1}{d_i} \]Solving for \( d_i \), we get:\[ \frac{1}{d_i} = \frac{1}{-12.0} - \frac{1}{20.0} = -\frac{1}{30.0} \]Thus, \( d_i = -30.0 \mathrm{cm} \). This means the image is 30.0 cm to the left of the first lens.

Step 2: Determine the Object Distance for the Converging Lens

The first image acts as the object for the second lens. The two lenses are 9.0 cm apart. The image from the first lens is located 30.0 cm to the left of it, i.e., 30.0 cm + 9.0 cm = 39.0 cm to the left of the second lens. Thus, the object distance for the second lens \( d'_o = -39.0 \mathrm{cm} \).

Step 3: Determine the Image Distance from the Converging Lens

For the second lens (a converging lens with \( f = 12.0 \mathrm{cm}\)), use the lens formula again:\[ \frac{1}{12.0} = \frac{1}{-39.0} + \frac{1}{d'_i} \]Solving for \( d'_i \), we get:\[ \frac{1}{d'_i} = \frac{1}{12.0} + \frac{1}{39.0} \approx \frac{1}{8.74} \]Thus, \( d'_i \approx 8.74 \mathrm{cm} \). The image is 8.74 cm to the right of the second lens or 17.74 cm (8.74 cm + 9.0 cm) to the right of the first lens.

Step 4: Check if the Final Image is Real or Virtual

The image distance from the second lens is positive (\( d'_i = 8.74 \mathrm{cm} \)), indicating the image is real, as real images are formed on the opposite side of the lens from the object.

Step 5: Calculate the Height of the Final Image

The magnification of the first lens is \( m_1 = -\frac{d_i}{d_o} = -\frac{-30.0}{20.0} = 1.5 \). The height of the first image is \( h'_i = 1.5 \times 2.50 \mathrm{mm} = 3.75 \mathrm{mm} \).The magnification of the second lens is \( m_2 = -\frac{d'_i}{d'_o} = -\frac{8.74}{-39.0} \approx 0.224 \). The final height is \( h''_i = m_1 \times m_2 \times h_o = 1.5 \times 0.224 \times 2.50 \mathrm{mm} \approx 0.84 \mathrm{mm} \).Since the final magnification is positive, the final image is erect.

Key Concepts

Thin Lens Formula

The thin lens formula is an essential tool in optics for finding the relationship between the object distance, image distance, and the focal length of a lens. The formula is given by:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]where:

• \( f \) is the focal length of the lens.
• \( d_o \) is the object distance, the distance between the object and the lens.
• \( d_i \) is the image distance, the distance between the image and the lens.

Understanding how to use this formula is crucial for solving problems involving lenses, as it allows one to determine the location where the image will be formed. Depending on the type of lens (converging or diverging), these values may be positive or negative, which helps identify the nature of the image (real or virtual). In the context of the exercise, this formula is used twice: once for the diverging lens and once for the converging lens. Each application helps in progressing from the position of the object to the final image location.

Image Distance

The image distance is the space between the lens and the projection of the image it creates. In lens combination problems, it often involves calculating the location of the image formed by more than one lens. For each lens, the image formed can act as the object for the next lens in the sequence.
In the exercise, we first find the image distance for the diverging lens using the thin lens formula, where the calculated image distance \( d_i \) is \(-30.0\) cm. A negative distance indicates the image is on the same side as the object, signifying a virtual image.
For the second lens, the converging one, the previous image acts as the new object. By substituting this into the thin lens formula once more, we determine that the image distance \( d'_i \) for the second lens is approximately \(8.74\) cm, on the opposite side from the incoming light, signifying a real image. Thus, understanding how to calculate the image distance allows us to determine not only the image's position but also its nature.

Focal Length

The focal length of a lens is a measure of how strongly it converges or diverges light. It is highly informative about the type of lens you are dealing with. In the context of the exercise:

• A positive focal length indicates a converging (or convex) lens which brings light to a focus.
• A negative focal length signifies a diverging (or concave) lens which spreads light out.

In our scenario, we have a diverging lens with a focal length of \(-12.0\) cm and a converging lens with a focal length of \(12.0\) cm. Knowing these values is crucial when applying the thin lens formula, as it affects how we calculate the image distances. The focal length not only helps in mathematical calculations but also gives insight into the physical behavior of the lens, guiding the solution process with clarity.

Magnification

Magnification tells us how much larger or smaller the image is compared to the object. It is calculated using:\[ m = -\frac{d_i}{d_o} \]In lens combinations, the total magnification is the product of the magnifications of each lens. Each lens affects the size and orientation of the image differently.For the first lens in our exercise, the magnification \( m_1 \) is \(1.5\), indicating the image is larger than the object. The negative sign would typically represent inversion, but since we multiply two negative values down the chain, it results in a positive total magnification, meaning the final image is erect.The final image's height is determined by multiplying the original height by the product of the magnifications of the lenses. This is what leads to the calculation of the final image being \(0.84\) mm tall. Understanding magnification gives us the final size and orientation of the image better in these compound lens arrangements.