Problem 63 - Physics

Question

A converging lens \((f=12.0 \mathrm{~cm})\) is located \(30.0 \mathrm{~cm}\) to the left of a diverging lens \((f=\) \(-6.00 \mathrm{~cm}\) ). A postage stamp is placed \(36.0 \mathrm{~cm}\) to the left of the converging lens. (a) Locate the final image of the stamp relative to the diverging lens. (b) Find the overall magnification, (c) Is the final image real or virtual? With respect to the original object, is the final image (d) upright or inverted, and is it (e) larger or smaller?

Step-by-Step Solution

Answer

The final image is 12 cm to the left of the diverging lens, virtual, upright, and smaller with a magnification of 0.5.

Step 1: Find the Image from the Converging Lens

We first find the image formed by the converging lens using the lens formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Here, \( f = 12.0 \mathrm{~cm} \) and \( d_o = 36.0 \mathrm{~cm} \). Substitute these values to find \( d_i \). \[ \frac{1}{12} = \frac{1}{36} + \frac{1}{d_i} \implies \frac{1}{d_i} = \frac{1}{12} - \frac{1}{36} \]\[ \frac{1}{d_i} = \frac{3 - 1}{36} = \frac{2}{36} = \frac{1}{18} \]Thus, \( d_i = 18 \mathrm{~cm} \). This means the image is 18 cm on the right of the converging lens.

Step 2: Determine Object Distance for the Diverging Lens

Now, use the image from the first lens as the object for the second lens. The distance between the lenses is 30 cm, so the object distance for the diverging lens is \( 30 \mathrm{~cm} - 18 \mathrm{~cm} = 12 \mathrm{~cm} \). Since the image is on the left of the diverging lens, the object distance \( d_o' = -12 \mathrm{~cm} \) (virtual object for the second lens).

Step 3: Find the Image from the Diverging Lens

Use the lens formula again for the diverging lens where \( f = -6.0 \mathrm{~cm} \) and \( d_o' = -12 \mathrm{~cm} \).\[ \frac{1}{-6} = \frac{1}{-12} + \frac{1}{d_i'} \]\[ \frac{1}{d_i'} = \frac{1}{-6} - \frac{1}{-12} = \frac{-2 + 1}{12} = -\frac{1}{12} \]Thus, \( d_i' = -12 \mathrm{~cm} \) which means the final image is virtual and 12 cm to the left of the diverging lens.

Step 4: Calculate the Overall Magnification

The magnification by each lens is given by \( m = -\frac{d_i}{d_o} \). First, find the magnification for the converging lens:\[ m_1 = -\frac{18}{36} = -0.5 \]Next, find the magnification for the diverging lens:\[ m_2 = -\frac{-12}{-12} = -1 \]The overall magnification is the product of the two:\[ m_{total} = m_1 \times m_2 = -0.5 \times (-1) = 0.5 \]

Step 5: Analyze the Final Image Characteristics

Based on the calculations: - **(c) Image is Virtual:** The final image is located on the same side as the object for the second lens, indicating it is virtual. - **(d) Image is Upright:** The overall magnification is positive, suggesting the image is upright. - **(e) Image is Smaller:** The magnification is less than 1, so the image is smaller than the object.

Key Concepts

Lenses

Lenses are transparent optical components, usually made of glass or plastic, that bend and focus light through refraction. They are fundamental tools in optics for image formation. There are two main types of lenses: converging (convex) lenses and diverging (concave) lenses.

• Converging Lenses: These lenses are thicker in the middle than at the edges and bring parallel light rays to a focus at a point known as the focal point. They have positive focal lengths.
• Diverging Lenses: These lenses are thinner in the middle and spread parallel light rays outward. Such lenses have negative focal lengths.

Lenses are characterized by their focal length, which is the distance from the lens to the focal point. The behavior of light through lenses is governed by the lens formula: \(rac{1}{f} = rac{1}{d_o} + rac{1}{d_i}\), where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. Understanding how light interacts with these lenses allows us to manipulate images in various optical devices like cameras, telescopes, and glasses.

Image Formation

The process of image formation involves the light rays converging or diverging as they pass through a lens to form an image. This image can be real or virtual, depending on the type of lens and the position of the object relative to the lens.

• Converging Lenses: When the object is placed outside the focal point, the rays converge to form a real image on the opposite side of the lens. The image can be projected onto a screen.
• Diverging Lenses: These always produce virtual images as the rays appear to diverge from a point behind the lens.

For image formation, the lens formula plays a crucial role. By plugging in the known values (focal length and object distance), you can calculate the image distance. This information tells you where the image will form in relation to the lens.

Magnification

Magnification is a measure of how much larger or smaller an image is compared to the object itself. It is defined by the ratio of the image height to the object height, and is also directly related to the distances involved in image formation. The formula for magnification in terms of distances is \( m = -\frac{d_i}{d_o}\), where \(m\) is the magnification, \(d_i\) is the image distance, and \(d_o\) is the object distance.

A positive magnification value indicates that the image is upright relative to the object, while a negative value suggests it is inverted. In practical terms:

• Greater than 1: The image is larger than the object, indicating magnification.
• Less than 1: The image is smaller than the object, indicating minification.

In optical systems involving multiple lenses, the total magnification is the product of the magnifications of each lens. This comprehensive understanding of magnification is vital for designing systems like microscopes and telescopes.

Virtual Image

A virtual image is formed when the light rays diverge, and thus cannot be projected onto a screen since they do not actually converge. Instead, the brain interprets these diverging rays as emanating from a location behind the lens, creating the illusion of an image. Virtual images possess unique properties:

• They are always upright. In contrast to real images, which can be inverted depending on the lens system.
• They cannot be captured on a screen because the light does not actually meet at the image position.
• They are commonly produced by diverging lenses or when an object is within the focal length of a converging lens.

Virtual images play a significant role in everyday optical instruments, like magnifying glasses and corrective lenses, helping users see an enlarged or clearer image when looking directly through the lens.