Problem 49 - Physics

Question

A diverging lens has a focal length of \(-25 \mathrm{~cm} .\) (a) Find the image distance when an object is placed \(38 \mathrm{~cm}\) from the lens. (b) Is the image real or virtual?

Step-by-Step Solution

Answer

The image distance is approximately \(-73.08 \mathrm{~cm}\) and the image is virtual.

Step 1: Understand the Lens Formula

The lens formula relates the object distance \(d_o\), the image distance \(d_i\), and the focal length \(f\) of the lens. It is given by \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). We will use this formula to find the image distance \(d_i\).

Step 2: Plug in the Known Values

From the problem, the object distance \(d_o = 38 \, \text{cm}\) and the focal length \(f = -25 \,\text{cm}\) for the diverging lens. We plug these values into the lens formula: \[ \frac{1}{-25} = \frac{1}{38} + \frac{1}{d_i}. \]

Step 3: Rearrange and Solve for Image Distance

First, solve the equation \(\frac{1}{d_i} = \frac{1}{-25} - \frac{1}{38}\). Find a common denominator (in this case, 950), then compute:\[ \frac{1}{d_i} = \frac{-38 + 25}{950} = \frac{-13}{950}. \]Thus, \(d_i = \frac{950}{-13} \approx -73.08 \,\text{cm}.\)

Step 4: Determine the Nature of the Image

For diverging lenses, if the image distance \(d_i\) is negative, the image is virtual. Since \(d_i \approx -73.08 \,\text{cm}\), the image is virtual.

Key Concepts

Lens Formula

The lens formula is an essential equation in optics, which helps us find the relationship between the object distance, the image distance, and the focal length of a lens. This 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 distance from the object to the lens,
• \( d_i \) is the distance from the image to the lens.

For a diverging lens, the focal length \( f \) is negative, which influences how light rays are bent by the lens. By using the lens formula, we can calculate unknown values when the other two are known. Plug in the known values and re-arrange the equation to solve for the unknown. This formula is a handy tool in predicting how lenses will affect the path of light rays, helping us understand the behavior of both real and virtual images.

Focal Length

Focal length is a crucial concept when dealing with lenses, as it describes how strongly a lens converges or diverges light. It is the distance between the lens and the point where parallel rays of light converge (for converging lenses) or appear to diverge from (for diverging lenses).

For diverging lenses, the focal length is negative. This negative sign indicates that the lens spreads out light rays, causing them to appear as though they are coming from a virtual point on the same side of the lens as the object. The magnitude of focal length indicates the lens's ability to bend light: shorter focal lengths indicate stronger bending, resulting in more divergence or convergence of light rays.

The focal length of lenses determines how large or small an image will be compared to the actual object and is fundamental in designing optical systems like glasses, cameras, and telescopes.

Image Distance

Image distance \( d_i \) is the distance from the image formed by a lens to the lens itself. It can be positive or negative, and this sign provides valuable information about the type of image produced. In the context of lenses, especially diverging lenses, understanding image distance helps in distinguishing between real and virtual images.

For a diverging lens, image distance often turns out to be negative based on the calculation from the lens formula. This indicates a virtual image, meaning the image cannot be projected onto a screen, as it forms on the same side of the lens as the object. Such images are often diminished in size compared to the object.

Knowing the image distance allows us to better understand the image's characteristics, such as whether it's enlarged, reduced, real, or virtual, and provides insights into the design and functionality of lens-based optical instruments.