Problem 97 - Fundamentals of Physics

Question

stick figure \(O\) (the object) stands on the common central axis of three thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closest to \(O\), which is at object distance \(p_{1} .\) Lens 2 is mounted within the middle boxed region, at distance \(d_{12}\) from lens \(1 .\) Lens 3 is mounted in the farthest boxed region, at distance \(d_{23}\) from lens \(2 .\) Each problem in Table \(34-10\) refers to a different combination of lenses and different values for distances, which are given in centimeters. The type of lens is indicated by \(C\) for converging and \(D\) for diverging: the number after \(\mathrm{C}\) or \(\mathrm{D}\) is the distance between a lens and either of the focal points (the proper sign of the focal distance is not indicated). Find (a) the image distance \(i_{2}\) for the (final) image produced by lens 3 (the final image produced by the system) and (b) the overall lateral magnification \(M\) for the system, including signs. Also, determine whether the final image is (c) real (R) or virtual (V), (d) inverted (I) from object \(O\) or noninverted (NI), and (e) on the same side of lens 3 as object \(O\) or on the opposite side. \(\begin{array}{lllllll}97 & +18 & \mathrm{C}, 6.0 & 15 & \mathrm{C}, 3.0 & 11 & \mathrm{C}, 3.0\end{array}\)

Step-by-Step Solution

Answer

(a) The final image distance \(i_2 = 7.5\) cm. (b) The overall magnification \(M = -0.75\). (c) The image is real (R). (d) The image is inverted (I). (e) The image is on the opposite side of lens 3.

Step 1: Understand Lens Setup

We have three lenses. Lens 1 is a converging lens (C) with a focal length of 6 cm. Lens 2 is a converging lens (C) with a focal length of 3 cm. Lens 3 is also a converging lens (C) with a focal length of 3 cm. The object distance for lens 1 is 18 cm, the distance between lens 1 and lens 2 is 15 cm, and the distance between lens 2 and lens 3 is 11 cm.

Step 2: Calculate Image Distance for Lens 1

Use the lens formula \(\frac{1}{f} = \frac{1}{p} + \frac{1}{i}\) to find the image distance \(i_1\) from lens 1. \(f_1 = 6\) cm, \(p_1 = 18\) cm. Therefore, \(\frac{1}{i_1} = \frac{1}{6} - \frac{1}{18}\). Solve for \(i_1\), which yields \(i_1 = 9\) cm.

Step 3: Determine Object Distance for Lens 2

The image from lens 1 becomes the object for lens 2. Calculate the object distance \(p_2\) for lens 2: \(p_2 = d_{12} - i_1 = 15 - 9 = 6\) cm.

Step 4: Calculate Image Distance for Lens 2

Apply the lens formula to lens 2. \(f_2 = 3\) cm, \(p_2 = 6\) cm. Find \(\frac{1}{i_2} = \frac{1}{3} - \frac{1}{6}\). Solving gives \(i_2 = 6\) cm.

Step 5: Determine Object Distance for Lens 3

The image from lens 2 acts as the object for lens 3. Determine the object distance \(p_3\) for lens 3: \(p_3 = d_{23} - i_2 = 11 - 6 = 5\) cm.

Step 6: Calculate Image Distance for Lens 3

Use the lens formula for lens 3. \(f_3 = 3\) cm, \(p_3 = 5\) cm. Find \(\frac{1}{i_3} = \frac{1}{3} - \frac{1}{5}\). Solving gives \(i_3 = 7.5\) cm. This is the final image distance \(i_2\) (part a).

Step 7: Calculate Overall Magnification

Calculate the magnification for each lens: \(M_1 = -\frac{i_1}{p_1} = -\frac{9}{18} = -0.5\), \(M_2 = -\frac{i_2}{p_2} = -\frac{6}{6} = -1\), \(M_3 = -\frac{i_3}{p_3} = -\frac{7.5}{5} = -1.5\). The overall lateral magnification \(M = M_1 \times M_2 \times M_3 = -0.5 \times -1 \times -1.5 = -0.75\) (part b).

Step 8: Evaluate Image Characteristics

C (c): Since the image distance \(i_3\) is positive, the final image is real (R). D (d): The magnification is negative, indicating that the image is inverted (I) compared to object O. E (e): Since the image is real (positive image distance), it is located on the opposite side of lens 3 relative to object O.

Key Concepts

Thin Lens Equation

The thin lens equation is a fundamental formula in geometric optics that relates the focal length of a lens, the object distance, and the image distance. This equation is expressed as: \[\frac{1}{f} = \frac{1}{p} + \frac{1}{i}\]where \( f \) is the focal length of the lens, \( p \) is the object distance, and \( i \) is the image distance. This formula enables us to determine the position where an image forms, given the location of the object and the lens characteristics.
For thin lenses, which are typically the focus in basic optics courses, the formula assumes the lens thickness is negligible compared to the object and image distances. Practical use of this equation involves rearranging it to solve for the desired unknown variable, like the image distance, which is crucial for understanding how lenses form images and affect image properties.

Image Distance Calculation

To compute the image distance, the thin lens equation is rearranged to solve for \( i \) (the image distance), once the object distance \( p \) and the focal length \( f \) are known. This is crucial in determining where the image will appear when light passes through a lens.
In the exercise, for three sequential lenses, these calculations were undertaken:

• For Lens 1: The object distance was 18 cm, and the focal length was 6 cm. Solving \( \frac{1}{i_1} = \frac{1}{6} - \frac{1}{18} \) gave an image distance \( i_1 \) of 9 cm.
• For Lens 2: The computed object distance from the image of Lens 1 and the focal length was used to determine \( i_2 \).
• For Lens 3: Finally, calculating \( i_3 \) using the lens formula, given the preceding values, resulted in an image distance of 7.5 cm.

Each lens affects the image location, effectively becoming the object for the subsequent lens in systems involving multiple lenses.

Lateral Magnification

Lateral magnification, often just called magnification, describes how much larger or smaller the image is compared to the object. It's calculated using the formula:\[M = -\frac{i}{p}\]where \( i \) is the image distance and \( p \) is the object distance. The negative sign indicates image inversion, meaning if the magnification is negative, the image is inverted relative to the object.

• For example, Lens 1 had \( M_1 = -\frac{9}{18} = -0.5 \), suggesting the image is half the size and inverted.
• Lens 2 yielded a magnification of -1, indicating a real image that is the same size but inverted.
• Lens 3's magnification, \( M_3 = -1.5 \), suggested the image is 1.5 times larger and inverted relative to the object.

Lateral magnification for a system of lenses is the product of each individual lens's magnification, resulting in an overall value that gives insight into the final image scale and orientation.

Image Characteristics

Image characteristics describe properties of the image formed by lenses, such as whether the image is real or virtual, inverted or non-inverted, and its position relative to the lens.
Based on the exercise:

• A **real image** occurs when the image distance \( i \) is positive, meaning light converges to form the image on the opposite side of the lens from the object.
• An **inverted image** results when the magnification is negative, indicating the image is flipped upside-down compared to the object.
• The final image is situated on the opposite side of Lens 3 from the object O due to the positive image distance \( i_3 \).

Understanding these characteristics is vital in optic systems design and analysis, where predicting the behavior of light and images is crucial for accurate, real-world application.