Problem 99 - University Physics with Modern Physics

Question

When an object is placed at the proper distance to the left of a converging lens, the image is focused on a screen 30.0 \(\mathrm{cm}\) to the right of the lens. A diverging lens is now placed 15.0 \(\mathrm{cm}\) to the right of the converging lens, and it is found that the screen must be moved 19.2 \(\mathrm{cm}\) farther to the right to obtain a sharp image. What is the focal length of the diverging lens?

Step-by-Step Solution

Answer

The focal length of the diverging lens is approximately -10.4 cm.

Step 1: Understanding the setup

First, identify the given distances. A converging lens focuses the object on a screen that is 30.0 cm to the right. When a diverging lens is added 15.0 cm to the right of the converging lens, the screen needs to be moved 19.2 cm further to the right for a clear image.

Step 2: Calculate final image distance

When the screen is moved, the new position from the converging lens to the screen is the original 30.0 cm plus the additional 19.2 cm, totaling 49.2 cm.

Step 3: Determine the virtual image location from the diverging lens

Since the diverging lens is 15.0 cm to the right of the converging lens, the distance from the diverging lens to the new image position (screen) is 49.2 cm - 15.0 cm = 34.2 cm.

Step 4: Use lens formula for the diverging lens

The lens formula is \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \). Substitute \( v = -34.2 \) cm (since the image is virtual) and \( u = 15.0 \) cm (distance between lenses). We solve for \( f \), the focal length of the diverging lens.

Step 5: Solve for focal length

Substituting the values into the lens equation: \[ \frac{1}{f} = \frac{1}{-34.2} - \frac{1}{15} \]Now calculate:\[ = -0.0292 - 0.0667 = -0.0959 \]Thus, the focal length \( f \) is \( \frac{1}{-0.0959} \approx -10.4 \) cm.

Key Concepts

Converging lens

A converging lens, sometimes known as a convex lens, is designed to bend light rays towards each other. This happens because the lens is thicker in the center than at the edges. Converging lenses have the special ability to focus parallel rays of light to a point called the "focal point."

• Used commonly for magnifying glasses, eyeglasses, cameras, and projectors.
• Known for forming real, inverted images if the object is placed outside the focal length.

When an object is positioned at the correct distance from a converging lens, it can project a sharply focused image on a screen. The distance from the lens to the image on the screen is known as the image distance. In the exercise, this setup initially forms an image 30 cm away from the lens.

Diverging lens

A diverging lens, also known as a concave lens, has the opposite effect of a converging lens. It is thinner in the center and causes light rays to spread apart. As a result, they never meet to form a real image but instead create what looks like an image at a point where the light doesn't actually reach. This is called a virtual image.

• Used in devices like peepholes, projectors, and beam expanders.
• It forms virtual, upright images regardless of the object distance.

In the given exercise, a diverging lens is placed 15 cm to the right of the converging lens, altering the image formation process. The screen had to be moved farther to create a sharp image again, indicating that the diverging lens changed the effective focal length of the system.

Lens formula

The lens formula is a fundamental equation used in optics to relate the object distance (\( u \), distance from the object to the lens), the image distance (\( v \), distance from the lens to the image), and the focal length (\( f \)) of the lens. The formula is:\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\]This formula is essential in determining focal length when object and image distances are known. The sign conventions are important:

• Positive image distance (\( v \)): real images formed on the opposite side of the light source.
• Negative image distance (\( v \)): virtual images appearing on the same side as the light source.
• Positive object distance (\( u \)) typically means the object is on the opposite side of the light direction.

For a lens system, like in the given exercise, the formula helps assess how lenses like the diverging lens alter the effective image and object distances.

Focal length calculation

Calculating the focal length of a lens involves applying the lens formula. In the exercise, we focused on finding the focal length of a diverging lens. This was achieved by understanding how the light refracted differently through two lenses and affected the image position. To find the focal length of the diverging lens, given the new image position and object distance:

• Image distance (\( v \)) from the diverging lens was found to be -34.2 cm. The negative sign indicates a virtual image formation.
• Object distance (\( u \)) was 15 cm, which is the distance between the lenses.

Applying these into the lens formula:\[\frac{1}{f} = \frac{1}{-34.2} - \frac{1}{15}\]Solving gives a focal length \( f \) of approximately -10.4 cm, which confirms the diverging nature of the lens. This negative value denotes that the lens spreads light, confirming it's a diverging lens.