Problem 24 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A diverging lens with a focal length of \(-14 \mathrm{~cm}\) is placed \(12 \mathrm{~cm}\) to the right of a converging lens with a focal length of \(18 \mathrm{~cm}\). An object is placed \(33 \mathrm{~cm}\) to the left of the converging lens. (a) Where will the final image be located? (b) Where will the image be if the diverging lens is \(38 \mathrm{~cm}\) from the converging lens?

Step-by-Step Solution

Answer

For (a), the final image is 9.09 cm left of the diverging lens. For (b), it's 12.57 cm left of the diverging lens.

Step 1: Find Image from Converging Lens

Using the lens formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), solve for the image distance \( d_i \) of the converging lens. Here, \( f = 18 \, \text{cm} \) and \( d_o = 33 \, \text{cm} \).The lens formula rearranges to:\[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{18} - \frac{1}{33} \]Calculating the values:\[ \frac{1}{d_i} = \frac{11 - 6}{198} = \frac{5}{198} \]Thus, \( d_i = \frac{198}{5} = 39.6 \, \text{cm} \). The image is formed 39.6 cm to the right of the converging lens.

Step 2: Analyze Situation for Diverging Lens Position at 12 cm

When the diverging lens is 12 cm to the right of the converging lens, the distance from the image of the converging lens to the diverging lens is \( |39.6 - 12| = 27.6 \, \text{cm} \). Thus, \( d_o' = 27.6 \, \text{cm} \) for the diverging lens.

Step 3: Find Image from Diverging Lens (12 cm Case)

Using the lens formula \( \frac{1}{f} = \frac{1}{d_o'} + \frac{1}{d_i'} \) for the diverging lens, where \( f = -14 \, \text{cm} \), solve for \( d_i' \).\[ \frac{1}{d_i'} = \frac{1}{-14} - \frac{1}{27.6} \]Calculate:\[ \frac{1}{d_i'} = \frac{-1}{14} - \frac{1}{27.6} \approx \frac{-27.6 - 14}{386.4} \]\[ \frac{1}{d_i'} \approx -0.110 \Rightarrow d_i' \approx -9.09 \, \text{cm} \]The negative sign indicates the image is 9.09 cm to the left of the diverging lens.

Step 4: Analyze Situation for Diverging Lens Position at 38 cm

When the diverging lens is 38 cm to the right of the converging lens, the distance from the image of the converging lens to the diverging lens is \( |39.6 - 38| = 1.6 \, \text{cm} \). Thus, \( d_o'' = 1.6 \, \text{cm} \) for the diverging lens.

Step 5: Find Image from Diverging Lens (38 cm Case)

Using the lens formula for the diverging lens again, where \( f = -14 \, \text{cm} \), solve for \( d_i'' \).\[ \frac{1}{d_i''} = \frac{1}{-14} - \frac{1}{1.6} \]Calculate:\[ \frac{1}{d_i''} = \frac{-1}{14} - \frac{1}{1.6} \approx \frac{-1 - 8.75}{22.4} \approx -0.4375 \]\[ d_i'' \approx -12.57 \, \text{cm} \]The image is approximately 12.57 cm to the left of the diverging lens.

Key Concepts

Diverging Lens

A diverging lens is specially designed to spread out light rays that pass through it. This spreading effect causes parallel rays of light to diverge as they exit the lens, which makes it ideal for applications where light needs to be dispersed.

• These lenses have a characteristic concave shape, often resembling a thin center with thicker edges.
• Light deviates outwardly, tracing back to a point called the focal point on the same side as the incoming light. This is known as the virtual focal point.
• The focal length of a diverging lens is negative, indicating its unique property of moving the light's convergence point to the opposite side of the lens.

In exercises like this one, a diverging lens affects the final position of the image by finalizing the outward direction of light rays that may have been directed by a converging lens first. Understanding the interaction between the focal point and light behavior is crucial in solving problems involving these lenses.

Converging Lens

Converging lenses work by bending incoming light rays towards a single point, known as the focal point. These lenses are typically convex in shape, appearing thicker in the center and thinner on the edges.

• They gather light to a focal point, providing a real focal length that is always positive.
• The practical use of converging lenses is seen in magnifying glasses and optical instruments, where bringing light to focus is essential.
• These lenses can produce both real and virtual images, depending on the position of the object relative to the lens.

In the context of the given exercise, the converging lens initially forms an image by directing light rays to converge at a particular point, and we calculate this using the lens formula. The interaction with a diverging lens then modifies this image location.

Lens Formula

The lens formula is a critical mathematical equation used to determine the relationships between the focal length, the distance of the object from the lens, and the distance of the image. The formula is written as:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]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.

This equation is fundamental in optics, allowing calculations of image position based on the known values of other variables.
In practical applications, such as the given exercise, the lens formula helps solve where the final image will be formed after passing through multiple lenses. By considering separately the impact of each lens and their respective focal lengths, understanding this equation is key to handling complex lens systems efficiently.