Problem 119 - Fundamentals of Physics

Question

A peanut is placed \(40 \mathrm{~cm}\) in front of a two-lens system: lens 1 (nearer the peanut) has focal length \(f_{1}=+20 \mathrm{~cm}\), lens 2 has \(f_{2}=-15 \mathrm{~cm}\), and the lens separation is \(d=10 \mathrm{~cm} .\) For the image produced by lens 2, what are (a) the image distance \(i_{2}\) (including sign), (b) the image orientation (inverted relative to the peanut or not inverted), and (c) the image type (real or virtual)? (d) What is the net lateral magnification?

Step-by-Step Solution

Answer

(a) The image distance \(i_2\) is calculated. (b) Its orientation is determined by net magnification sign. (c) Type is based on \(i_2\) sign. (d) Magnification is product of individual magnifications.

Step 1: Locate the image formed by lens 1

First, calculate the image distance from lens 1 using the lens formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) where \( f_1 = +20 \) cm and \( d_o = 40 \) cm. Plug these values into the formula: \( \frac{1}{20} = \frac{1}{40} + \frac{1}{d_{i1}} \). Solve for \( d_{i1} \) to find the image distance from lens 1.

Step 2: Translate image distance to lens 2

The distance of the image from lens 1 will act as the object for lens 2. Calculate the new object distance from lens 2: \( d_{o2} = d_{i1} - d \), where \( d \) is the lens separation of 10 cm.

Step 3: Locate the image formed by lens 2

Use the lens formula again for lens 2: \( \frac{1}{f_2} = \frac{1}{d_{o2}} + \frac{1}{d_{i2}} \), where \( f_2 = -15 \) cm. Substitute \( d_{o2} \) from Step 2 into this equation and solve for \( d_{i2} \) to find the image distance from lens 2.

Step 4: Determine image orientation and type

Calculate the magnifications. For lens 1: \( m_1 = -\frac{d_{i1}}{d_o} \), and for lens 2: \( m_2 = -\frac{d_{i2}}{d_{o2}} \). If the net magnification, \( m = m_1 \times m_2 \), is negative, the image is inverted; if positive, it is upright. Real images are formed when \( d_{i2} \) is positive, and virtual images are formed when \( d_{i2} \) is negative.

Step 5: Calculate net lateral magnification

Combine the magnification factors to find the net lateral magnification: \( m = m_1 \times m_2 \). Substitute the magnifications calculated in Step 4 and solve for the net lateral magnification.

Key Concepts

Optical Lenses

Optical lenses are transparent pieces of material, typically glass or plastic, that refract light in such a way as to converge or diverge it. Lenses have two sides, each of which can be convex (curved outward) or concave (curved inward). The shape of the lens affects how light rays pass through it and how an image is formed on the other side.
There are two main types of lenses:

• **Convex Lenses:** These lenses are thicker at the middle. They cause parallel rays of light to converge to a focal point. Convex lenses are often used in magnifying glasses and cameras.
• **Concave Lenses:** These lenses are thinner in the middle. They cause parallel rays of light to spread out as if they had originated from a focal point behind the lens. Concave lenses are commonly found in eyeglasses for correcting nearsightedness.

Understanding how lenses work is crucial for many applications, including cameras, microscopes, and eyeglasses, which rely on the precise control of light and image formation.

Image Formation

Image formation by a lens involves bending light rays and bringing them together to create a likeness of the object. The process differs depending on whether the lens is convex or concave.
When light rays pass through a:

• **Convex Lens:** Rays converge at a single focal point, creating a real image. This real image can be projected on a screen and is inverted compared to the object.
• **Concave Lens:** Rays diverge, and the image appears to originate from a point behind the lens, resulting in a virtual image that cannot be projected on a screen and is upright.

In a lens system consisting of two lenses, like the problem states, the image from the first lens acts as the object for the second lens. This can result in complex transformations leading to various combinations of real and virtual, inverted or upright images.

Focal Length

The focal length of a lens is the distance from the center of the lens to the focal point, where like rays converge or appear to diverge. It is a crucial parameter that influences how a lens focuses light and forms an image.
A few points to keep in mind:

• **Positive Focal Length:** Indicates a converging (convex) lens, focusing light to a point, forming real and inverted images.
• **Negative Focal Length:** Refers to a diverging (concave) lens that spreads light out, resulting in virtual and upright images.

In the exercise, lens 1 with a focal length of +20 cm is a convex lens, while lens 2 with a focal length of -15 cm is concave. The calculations for image formation involve using the lens formula:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]This relationship helps in determining where the image will form relative to the lens.

Magnification

Magnification in optics indicates how much larger or smaller an image is compared to the object. It's a dimensionless number that helps understand the scale of the image produced by the lens.
The magnification formula is given as:

• **For a Single Lens:** \( m = -\frac{d_i}{d_o} \), where \( d_i \) is the image distance and \( d_o \) is the object distance. The negative sign indicates that an image inversion has occurred.
• **For a System of Lenses:** The net magnification is the product of the magnifications of individual lenses, \( m = m_1 \times m_2 \). A negative result indicates an inverted image, while a positive result indicates an upright image.

Magnification also tells us about the types of images. If the overall value is less than 1, the image is diminished; if greater than 1, it is magnified. Calculating the magnification helps in predicting how the final image will appear when using multiple lenses, and it's crucial for devices like telescopes or microscopes, where precise image size adjustments are necessary.