Problem 51 - College Physics

Question

A diverging lens with a focal length of \(-48.0 \mathrm{cm}\) forms a virtual image 8.00 \(\mathrm{mm}\) tall, 17.0 \(\mathrm{cm}\) to the right of the lens. (a) Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? (b) Draw a principal-ray diagram for this situation.

Step-by-Step Solution

Answer

(a) The object is 26.32 cm from the lens, 1.24 cm tall, and the image is inverted. Object and image are on the same side. (b) Draw the principal-ray diagram based on given steps.

Step 1: Understand the Given Information

We are given a diverging lens with a focal length \( f = -48.0 \text{ cm} \). The image is virtual, \( h' = 8.00 \text{ mm} = 0.8 \text{ cm} \) tall, and located \( 17.0 \text{ cm} \) to the right of the lens. We need to determine the object's position \((d_o)\) and size \((h_o)\), as well as the nature of the image. We'll use the lens formula and magnification formula to solve this.

Step 2: Use the Lens Formula

The lens formula is given by \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Here, \( f = -48.0 \text{ cm} \) and \( d_i = -17.0 \text{ cm} \) (negative because the image is virtual). Substitute these values into the formula to find \(d_o\).\[ \frac{1}{-48.0} = \frac{1}{d_o} - \frac{1}{17.0} \]Solve for \( \frac{1}{d_o} \) to find the object distance \(d_o\).

Step 3: Solve for Object Distance \(d_o\)

Rearrange the lens equation to solve for \(\frac{1}{d_o}\).\[ \frac{1}{d_o} = \frac{1}{-48.0} + \frac{1}{17.0} \]Calculate:\[ \frac{1}{d_o} = \frac{-1}{48.0} + \frac{1}{17.0} \]Put both over a common denominator to find \(d_o\).

Step 4: Calculate the Object Distance \(d_o\)

Calculate the combined fractions from Step 3:\[ \frac{-1}{48.0} + \frac{1}{17.0} = \frac{-17 + 48}{816} = \frac{31}{816} \]Thus, \(d_o = \frac{816}{31} \approx 26.32 \text{ cm}\). So, the object is approximately 26.32 cm from the lens.

Step 5: Determine the Magnification and Object Height

The magnification \( m \) is given by \( m = \frac{h'}{h_o} = \frac{d_i}{d_o} \). We know \(d_i = -17.0\) and the height of the image \(h' = 0.8 \text{ cm}\).First calculate \(m\) using magnifying formula:\[ m = \frac{-17.0}{26.32} = -0.645 \]Then use \(m = \frac{h'}{h_o}\) to find the object height \(h_o\):\[ h_o = \frac{h'}{m} = \frac{0.8}{-0.645} \approx -1.24 \text{ cm} \].

Step 6: Analyze the Image Characteristics

Since the image distance \(d_i\) is negative, it confirms the image is virtual and on the same side of the lens as the object. The negative magnification \(m\) indicates the image is inverted relative to the object.

Step 7: Draw a Principal-Ray Diagram (Optional)

To visualize, draw a simple sketch: - Mark the diverging lens. - Draw an object arrow on the left. - Show principal rays: one parallel, then diverging through the focal point; one through the center of the lens continuing straight. - The virtual image appears on the same side as the object, confirming the calculations.

Key Concepts

lens formulas

In optics, lens formulas are fundamental tools used to determine the relationship between the object distance, image distance, and the focal length of a lens. The most commonly used formula is the lens equation, which is:
\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Here:

• \(f\) is the focal length of the lens.
• \(d_o\) is the distance from the object to the lens.
• \(d_i\) is the distance from the image to the lens.

When dealing with diverging lenses, the focal length \(f\) is always negative, reflecting the lens's ability to spread out light rays. By plugging in known values into this formula, one can solve for the unknowns, as demonstrated in our original exercise where the object distance \(d_o\) was calculated. The lens formula is an essential part of solving numerous optics problems, allowing for the determination of image characteristics such as position and orientation.

virtual image

In optics, a virtual image is an image formed when outgoing rays from a point on an object do not converge in reality, but appear to do so when extended backward. Virtual images are upright with respect to the object when viewed through lenses or mirrors. In the case of a diverging lens, the virtual image appears on the same side as the object and is always formed by light rays that do not actually converge but appear to diverge from a common point.
In the exercise, the virtual image was indicated by a negative image distance \(d_i = -17.0 \text{cm}\), consistent with the conventions used for optical calculations. Virtual images are typically smaller or magnified depending on the lens's properties and the object's proximity to the lens.

image magnification

Magnification, in the context of optics, refers to the process of enlarging or reducing the appearance of an object through a lens. The magnification formula is given by:
\[ m = \frac{h'}{h_o} = \frac{d_i}{d_o} \]where:

• \(m\) is the magnification factor.
• \(h'\) is the height of the image.
• \(h_o\) is the height of the object.
• \(d_i\) is the image distance.
• \(d_o\) is the object distance.

In the example provided, a negative magnification was calculated, which confirmed that the image was inverted compared to the object. Magnification helps us quantify the change in size and orientation of the image produced by a lens, which is crucial in applications ranging from simple magnifying glasses to complex scientific instruments.

optics

Optics is a branch of physics that studies the behavior and properties of light and its interactions with matter. Among the many focuses of optics is the study of lenses, which are devices that refract light in a way that forms images of objects. Lenses are typically classified as either converging or diverging, based on how they bend light rays.
A diverging lens, like the one described in the exercise, spreads out parallel light rays as if they were emanating from a common focal point. Such lenses are often thinner in the middle and thicker at the edges. Studying the behavior of light through lenses allows us to understand and predict how images will form, involving practical applications in everything from eyeglasses to cameras.
Understanding optics is crucial for anyone learning about how light interacts with lenses and how images are produced and manipulated, making it an integral part of fields related to vision technology, photography, and optical instruments.