Problem 25 - Physics for Scientists and Engineers with Modern Physics

Question

(II) Two lenses, one converging with focal length \(20.0 \mathrm{~cm}\) and one diverging with focal length \(-10.0 \mathrm{~cm},\) are placed \(25.0 \mathrm{~cm}\) apart. An object is placed \(60.0 \mathrm{~cm}\) in front of the converging lens. Determine \((a)\) the position and \((b)\) the magnification of the final image formed. (c) Sketch a ray diagram for this system.

Step-by-Step Solution

Answer

(a) Final image at -3.33 cm (virtual). (b) Total magnification is -0.333. (c) Ray diagram shows image positions.

Step 1: Understand the Lens Equation

The lens equation is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. Our task involves using this formula twice: once for each lens.

Step 2: Calculate Image Position from the First Lens

For the converging lens, \( f = 20.0 \, \text{cm} \) and \( d_o = 60.0 \, \text{cm} \). Using the lens formula:\[\frac{1}{20} = \frac{1}{60} + \frac{1}{d_{i1}}\]Solving for \( d_{i1} \), we get \( d_{i1} = 30.0 \, \text{cm} \). So, the image from the first lens is formed 30.0 cm behind it.

Step 3: Determine Object Distance for the Second Lens

The image from the first lens acts as the object for the second lens. The separation between the lenses is 25.0 cm, so the object distance for the second lens \( d_{o2} = 30.0 \text{cm} - 25.0 \text{cm} = 5.0 \text{cm} \) (positive because the image is on the same side as the light coming towards the second lens).

Step 4: Calculate Image Position from the Second Lens

For the diverging lens, \( f = -10.0 \, \text{cm} \) and \( d_{o2} = 5.0 \, \text{cm} \). Using the lens formula:\[\frac{1}{-10} = \frac{1}{5} + \frac{1}{d_{i2}}\]Solving for \( d_{i2} \), we get \( d_{i2} = -3.33 \, \text{cm} \). This negative sign indicates the image is virtual and on the same side as the object for the second lens.

Step 5: Calculate the Magnification of the System

The magnification due to the first lens is \( m_1 = -\frac{d_{i1}}{d_{o1}} = -\frac{30}{60} = -0.5 \).The magnification due to the second lens is \( m_2 = -\frac{d_{i2}}{d_{o2}} = -\frac{-3.33}{5} = 0.666 \).The total magnification \( m = m_1 \times m_2 = (-0.5) \times (0.666) = -0.333 \).

Step 6: Drawing the Ray Diagram

Draw two lenses, the first being converging followed by a diverging one, with an object placed 60.0 cm in front of the first lens. Trace rays: 1. A ray parallel to the principal axis refracts through the focal point on the other side. 2. A ray through the center of the lens continues straight. Note the image formed by the first lens becomes the object for the second lens. Repeat the tracing for the second lens with the new object position.

Key Concepts

Converging Lenses

A **converging lens**, often called a convex lens, can focus parallel rays of light to a single point known as the focal point. This type of lens is thicker in the middle than at the edges, which helps bend the rays towards the focal point.

• In our exercise, the converging lens has a focal length of 20.0 cm.
• The object is placed 60.0 cm in front of it.

Using the lens formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), we calculate that the image formed is 30.0 cm behind the converging lens. This real image is created on the opposite side from where light enters the lens.
The lens formula is crucial for determining image positions in lens systems. It helps us understand how the converging lens transforms the object into an image, which then serves as the object for subsequent lenses in any system.

Diverging Lenses

A **diverging lens** is the opposite of a converging lens. It spreads out light rays that are initially parallel. Often, these lenses are shaped such that they are thinner in the middle than at the edges, resembling a concave shape. These lenses have a virtual focal point, as the rays appear to diverge from a single point on the same side as the light source.
For the diverging lens in the exercise, with a focal length of -10.0 cm:

• The new object distance for the diverging lens is 5.0 cm.
• The calculation gives an image distance of -3.33 cm, indicating the image is virtual.

This negative image distance informs us that the image does not actually form on a physical point but appears to emanate from there when traced backward. Diverging lenses thus form virtual images due to this separation of rays, crucial for corrective optics like glasses.

Ray Diagrams

**Ray diagrams** are visual tools used to predict the image formed by lenses. They illustrate how light rays interact with optical elements, like converging and diverging lenses in our exercise. Ray diagrams help visualize both real and virtual image formations.
To sketch a ray diagram:

• For a converging lens: draw a ray parallel to the principal axis, refracting through the focal point on the opposite side. Also, draw a ray through the lens center that continues straight.
• For a diverging lens: depict a ray appearing to emerge from the focal point even though it strikes the lens parallel to the principal axis. Another ray passing through the lens center also continues straight.

Observing these diagrams shows how the first lens's image acts as the input for the second lens. This understanding is key in multiple-lens systems, finding application in devices like microscopes and telescopes, where precise image positioning is critical.