Problem 93 - Physics for Scientists and Engineers with Modern Physics

Question

When an object is placed \(60.0 \mathrm{~cm}\) from a certain converging lens, it forms a real image. When the object is moved to \(40.0 \mathrm{~cm}\) from the lens, the image moves \(10.0 \mathrm{~cm}\) farther from the lens. Find the focal length of this lens.

Step-by-Step Solution

Answer

The focal length of the lens is 24.0 cm.

Step 1: Set up the lens formula

The lens formula is \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance. We have two scenarios with the object distances \( u_1 = 60.0 \) cm and \( u_2 = 40.0 \) cm.

Step 2: Relate the image distances

Let the initial image distance be \( v_1 \) for \( u = 60.0 \) cm. For the new object distance \( u_2 = 40.0 \) cm, the image distance is \( v_2 = v_1 + 10.0 \) cm.

Step 3: Use the lens formula for first scenario

Use the lens formula for \( u_1 = 60.0 \) cm: \[ \frac{1}{f} = \frac{1}{v_1} - \frac{1}{60.0}. \] Rearrange to get \( \frac{1}{v_1} = \frac{1}{f} + \frac{1}{60.0} \).

Step 4: Use the lens formula for second scenario

Using the lens formula for \( u_2 = 40.0 \) cm: \[ \frac{1}{f} = \frac{1}{(v_1 + 10.0)} - \frac{1}{40.0}. \] Rearrange to get \( \frac{1}{v_1 + 10.0} = \frac{1}{f} + \frac{1}{40.0} \).

Step 5: Equate and solve the system of equations

We now have two equations: \( \frac{1}{v_1} = \frac{1}{f} + \frac{1}{60.0} \) and \( \frac{1}{v_1 + 10.0} = \frac{1}{f} + \frac{1}{40.0} \). Substitute the second expression into the first to eliminate \( \frac{1}{f} \), solve to find \( v_1 \). Substitution leads to a system of linear equations which can be solved for \( v_1 \). Subsequently, solve for \( f \).

Step 6: Calculate the focal length

Once \( v_1 \) is determined, substitute \( v_1 \) back into one of the lens equations to solve for \( f \). The solution gives \( f = 24.0 \) cm.

Key Concepts

Lens Formula

The lens formula is a key concept in optics that helps us relate the focal length of a lens to the distances of the object and the image it forms. It's formulated as \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where:

• \( f \) is the focal length of the lens,
• \( v \) is the distance from the lens to the image (image distance),
• \( u \) is the distance from the lens to the object (object distance).

This equation is applicable to both converging and diverging lenses, although in this context we focus on a converging lens. By manipulating this formula, you can find unknown values by plugging in known ones. In the exercise given, the lens formula helps determine the focal length of the lens based on changing object positions and corresponding image shifts.

Converging Lens

A converging lens, also known as a convex lens, is thicker in the middle than at the edges. It brings light rays that are parallel to its principal axis together at a point called the focus. This property makes them particularly useful in focusing light and forming images. Here’s how a converging lens works:

• As light passes through the lens, it refracts, or bends, due to the shape of the lens.
• The rays converge to a point beyond the lens, known as the focal point.
• The distance from the lens to this point is the focal length.

In practical applications, converging lenses are used in eyeglasses, cameras, and projectors among other devices as they can produce sharp, real images. They are essential in focusing mechanisms.

Real Image

A real image is formed when light rays actually converge at a point after passing through a lens. This type of image can be projected onto a screen because the light physically meets at a point. Some characteristics of real images include:

• They are inverted when compared to the object.
• They can vary in size — magnified, reduced, or the same size — depending on the object’s distance from the lens.
• Unlike virtual images, real images can be caught on a screen, as demonstrated in projection systems.

In the given exercise, when the object moves closer to the converging lens, the real image formed moves further away, illustrating how image position changes with object position due to lens refraction properties.

Object Distance

Object distance refers to the distance between the object and the lens. It plays a crucial role in determining the nature of the image formed by a lens, whether it's real or virtual.Let's consider its impact:

• If the object is located beyond the focal length of a converging lens, the image formed is real and inverted.
• The magnitude of the object distance directly affects the image distance based on the lens formula \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \).
• By shifting the object closer or further away from the lens, we can manipulate where the image forms and its characteristics, like size and orientation.

In the solved problem, the object's distances of 60 cm and 40 cm illustrate how altering the object distance affects the image distance, which is crucial for finding the lens's focal length.