Problem 85 - Fundamentals of Physics

Question

Two-lens systems. In Fig. \(34-45\), stick figure \(O\) (the object) stands on the common central axis of two thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closer to \(O\), which is at object distance \(p_{1}\). Lens 2 is mounted within the farther boxed region, at distance \(d .\) Each problem in Table \(34-9\) refers to a different combination of lenses and different values for distances, which are given in centimeters. The type of lens is indicated by \(\mathrm{C}\) for converging and D for diverging; the number after C or \(\mathrm{D}\) is the distance between a lens and either of its focal points (the proper sign of the focal distance is not indicated). Find (a) the image distance \(i_{2}\) for the image produced by lens 2 (the final image produced by the system) and (b) the overall lateral magnification \(M\) for the system, including signs. Also, determine whether the final image is (c) real (R) or virtual (V), (d) inverted (I) from object \(O\) or noninverted (NI), and (e) on the same side of lens 2 as object \(O\) or on the opposite side. \(\begin{array}{lllll}\mathbf{8 5} & +4.0 & \text { C, } 6.0 & 8.0 & \text { D, } 6.0\end{array}\)

Step-by-Step Solution

Answer

The final image is virtual, non-inverted, and on the same side of lens 2 as the object.

Step 1: Analyze Given Information

We are given two lenses and their parameters: Lens 1 is converging with a focal length of 6.0 cm, and Lens 2 is diverging with a focal length of 6.0 cm. The object distance for Lens 1 is given as \(p_1 = 85\) cm, and the distance between the two lenses is \(d = 4.0\) cm.

Step 2: Calculate Image Position from Lens 1

For Lens 1, use the lens formula \(\frac{1}{f_1} = \frac{1}{p_1} + \frac{1}{i_1}\), where \(f_1 = 6.0\) cm. Solve for \(i_1\):\[\frac{1}{6.0} = \frac{1}{85} + \frac{1}{i_1}\]Solving gives \(i_1 = \frac{1}{\frac{1}{6} - \frac{1}{85}}\). Calculate \(i_1\) to find the intermediate image position.

Step 3: Check Intermediate Image Position

Calculate \(i_1\):\[\frac{1}{i_1} \approx \frac{1}{6} - \frac{1}{85} = \frac{85 - 6}{510} = \frac{79}{510}\]\[i_1 \approx \frac{510}{79} \approx 6.46\] cm. Since \(i_1\) is positive, the image is real and on the opposite side of Lens 1.

Step 4: Determine Object Distance for Lens 2

The object for Lens 2 is the image from Lens 1. Compute the object distance \(p_2\) for Lens 2:\[p_2 = d - i_1 = 4.0 - 6.46 = -2.46\] cm. The negative value indicates the object for Lens 2 is virtual and on the same side as its direction of incoming light.

Step 5: Calculate Image Position from Lens 2

Use the lens formula for Lens 2, \(\frac{1}{f_2} = \frac{1}{p_2} + \frac{1}{i_2}\), where \(f_2 = -6.0\) cm and \(p_2 = -2.46\) cm (diverging lens):\[\frac{1}{-6.0} = \frac{1}{-2.46} + \frac{1}{i_2}\]Solving gives \(i_2 = \frac{1}{\frac{1}{2.46} - \frac{1}{6}}\). Calculate \(i_2\) to find the final image position.

Step 6: Calculate final image position

Compute \(i_2\):\[\frac{1}{i_2} \approx \frac{-1}{2.46} - \frac{1}{6} = \frac{-6 - 2.46}{14.76} = \frac{-8.46}{14.76}\]\[i_2 \approx \frac{-14.76}{8.46} \approx -1.74\] cm. The negative sign indicates a virtual image, located on the same side as object's incoming light for Lens 2.

Step 7: Calculate Lateral Magnification

The overall magnification \(M\) is the product of the magnifications of each lens: \[M = m_1 \times m_2 = \left(-\frac{i_1}{p_1}\right) \times \left(-\frac{i_2}{p_2}\right)\]Calculate \(m_1 = -\frac{6.46}{85}\) and \(m_2 = -\frac{-1.74}{-2.46}\), then compute \(M = m_1 \times m_2\).

Step 8: Classification of Final Image

Using the results:- The image is virtual (negative \(i_2\)).- The image is non-inverted since both \(m_1\) and \(m_2\) are negative, making \(M\) positive.- The image is on the same side of Lens 2 as the object due to a negative \(i_2\).

Key Concepts

Converging Lens

A converging lens, also known as a convex lens, bends light rays toward each other, causing them to meet at a point. This focal point is crucial in forming images and is where the lens concentrates incoming light.

Converging lenses have several important properties:

• They are thicker in the middle than at the edges.
• They can produce both real and virtual images, depending on the object's position relative to the lens's focal point.
• When an object is placed further away than the focal length, the lens forms a real and inverted image.
• If the object is within the focal length, a virtual image appears, which is upright and magnified.

This characteristic allows converging lenses to be used in eyeglasses, cameras, and other optical devices that require focusing light.

Diverging Lens

A diverging lens, or concave lens, spreads light rays outward as if they originate from a point. This behavior defines their unique role in optics, particularly in creating virtual images.

Key features of diverging lenses include:

• They are thinner in the middle than at the edges.
• Only form virtual, upright, and reduced images regardless of object position.
• The virtual image forms on the same side as the incoming light, which can be useful in correcting vision issues like myopia.
• Diverging lenses have a negative focal length, indicating the direction of focal points opposite to converging lenses.

By controlling the spreading of light, diverging lenses find applications in devices like peepholes and corrective glasses.

Image Distance Calculation

Image distance calculation involves determining the distance between the lens and the point at which the image is formed. This provides key insights into the nature and characteristics of the image.To calculate image distance:

• Use the lens formula: \( \frac{1}{f} = \frac{1}{p} + \frac{1}{i} \), where \( f \) is the focal length, \( p \) is the object distance, and \( i \) is the image distance.
• Rearrange the equation to solve for \( i \): \( i = \frac{1}{\frac{1}{f} - \frac{1}{p}} \).
• Pay attention to the sign of \( i \), as positive indicates a real image, while negative signifies a virtual image.

Understanding this calculation helps in predicting the behavior and location of images formed by lens systems.

Lateral Magnification

Lateral magnification measures how much larger or smaller the image is compared to the object. It is a scaling factor that demonstrates the extent of enlargement or reduction by a lens system.To calculate lateral magnification:

• The formula used is \( M = m_1 \times m_2 = \left(-\frac{i_1}{p_1}\right) \times \left(-\frac{i_2}{p_2}\right) \), where \( i_1 \) and \( i_2 \) are image distances, \( p_1 \) and \( p_2 \) are object distances, and \( m_1 \) and \( m_2 \) are magnifications of each lens.
• A positive magnification indicates an upright image, while a negative value shows an inverted image.
• Lateral magnification tells us about the orientation (inverted or upright) and relative size of an image compared to the object.

Lateral magnification is vital for understanding image properties in optical systems like microscopes and telescopes.

Virtual Image

A virtual image is one that cannot be displayed on a screen as it appears to form on the same side of the lens as the object. Unlike real images that are produced by converging light rays, virtual images result from diverging rays.

Virtual images possess several distinct characteristics:

• They are typically upright and can appear larger or smaller than the object.
• Produced by diverging lenses or situations where the object is within the focal length of a converging lens.
• Are an essential feature of optical instruments such as magnifying glasses and rear-view mirrors.
• Require eye interpretation or additional lenses to be seen as they don't project onto a surface.

Understanding virtual images helps in grasping concepts behind optics that manipulate light paths for various applications.