Problem 62 - Physics

Question

At offers a review of the concepts that play roles in this problem. A diverging lens \((f=-10.0 \mathrm{~cm})\) is located \(20.0 \mathrm{~cm}\) to the left of a converging lens \((f=30.0 \mathrm{~cm})\). A \(3.00-\mathrm{cm}\) -tall object stands to the left of the diverging lens, exactly at its focal point. (a) Determine the distance of the final image relative to the converging lens. (b) What is the height of the final image (including the proper algebraic sign)?

Step-by-Step Solution

Answer

(a) 30.0 cm to the right; (b) 3.00 cm (upright).

Step 1: Understand the Setup

We have two lenses: a diverging lens with focal length \( f_1 = -10.0 \, \mathrm{cm} \) and a converging lens with \( f_2 = 30.0 \, \mathrm{cm} \). They are spaced \( 20.0 \, \mathrm{cm} \) apart. An object that is \( 3.00 \, \mathrm{cm} \) tall is placed at the focal point of the diverging lens, i.e., \( d_o = -10.0 \, \mathrm{cm} \) from the diverging lens.

Step 2: Determine Image Formed by Diverging Lens

For the diverging lens, use the lens formula: \[ \frac{1}{f_1} = \frac{1}{d_o} + \frac{1}{d_i} \] Here \( d_o = -10.0 \, \mathrm{cm} \) because the object is placed at the focal point.Substituting the values: \[ \frac{1}{-10.0} = \frac{1}{-10.0} + \frac{1}{d_i} \] Solving, we find \( d_i = \infty \), so the first image is formed at infinity.

Step 3: Recognize Parallel Light for Converging Lens

Since the image from the diverging lens is at infinity, light enters the converging lens parallel. Therefore, the converging lens forms its image at its focal point. Since the lenses are \( 20.0 \, \mathrm{cm} \) apart, the image distance \( d_i' \) from the converging lens is equal to its focal length \( f_2 = 30.0 \, \mathrm{cm} \), pointing to the right of the converging lens.

Step 4: Calculate Image Distance from Converging Lens

The image from the converging lens is formed at its focal point since the incoming light rays were parallel. Image distance: \[ d_i' = f_2 = 30.0 \, \mathrm{cm} \] Thus, the final image is \( 30.0 \, \mathrm{cm} \) to the right of the converging lens.

Step 5: Determine Height of Final Image

In this optical setup, the magnification for the diverging lens is 0 since its image appears infinitely far. The light is parallel to the converging lens, creating an image of the same height as the object. Therefore, the height of the final image remains \( h = 3.00 \, \mathrm{cm} \) tall. The image is not inverted (remains upright), so the height is positive.

Step 6: Compile the Results

(a) The distance of the final image relative to the converging lens is \( 30.0 \, \mathrm{cm} \) to the right. (b) The height of the final image is \( 3.00 \, \mathrm{cm} \) (positive, as it remains upright).

Key Concepts

Diverging Lens

A diverging lens is a type of optical lens that causes light rays to spread out or diverge. This lens is thinner at the center than at the edges. A common everyday example would be eyeglasses used for nearsightedness. When light rays pass through a diverging lens, they bend outward. This makes an object appear smaller and further away.

In an optical setup, such as the one described in the exercise, diverging lenses typically form virtual images. These virtual images can't be projected onto a screen since the light rays don't actually meet in reality. To determine the behavior of a diverging lens mathematically, you can use the lens formula, which relates object distance, image distance, and focal length.

Converging Lens

Unlike a diverging lens, a converging lens has thicker centers compared to its edges. This type of lens is often used in magnifying glasses or eyeglasses for farsighted people. It focuses incoming light rays to a single point known as the focal point.

In the exercise, when light coming from the diverging lens enters a converging lens, the rays are parallel and focus at the lens's focal point. Thus, the final image formed by the converging lens in our setup appears real and inverted. To calculate image distances or magnifications with converging lenses, scientists use the lens formula, similar to the diverging lens.

Image Distance

Image distance is a crucial concept when talking about lenses. It is defined as the distance between the lens and the image it forms. This value helps us determine where the image appears, which is essential in optical calculations and applications.

In our exercise, the diverging lens initially forms an image at infinity, which means the light appears parallel as it exits the lens. When these parallel rays enter the converging lens, they focus at that lens's focal point. Therefore, the image distance from the converging lens is equal to its focal length.

Focal Point

The focal point of a lens is a vital aspect in lens optics. It refers to the point where converging light rays meet or appear to meet after passing through a lens. Both diverging and converging lenses have a focal point, although their effects differ.

For the diverging lens in the exercise, the focal point indicates where the image would seem to originate. However, the light rays themselves diverge, never actually meeting. In contrast, for the converging lens, the focal point is where all parallel light rays are directed and focused. Knowing the position of the focal points helps determine how and where an image forms in complex setups.

Lens Formula

The lens formula is an equation that relates three main aspects of a lens's function: the focal length (\( f \)), the object distance (\( d_o \)), and the image distance (\( d_i \)). It is given by the equation: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]This formula is instrumental when solving problems related to lens optics.

In the exercise, it is used to calculate the distances at which images are formed. Though the focal lengths and the positions of the lenses are crucial, this formula helps to link and solve for unknown distances. By using it step by step, one can derive both where and how the image appears, be it real or virtual.