Problem 61 - Physics

Question

A converging lens \(\left(f_{1}=24.0 \mathrm{~cm}\right)\) is located \(56.0 \mathrm{~cm}\) to the left of a diverging lens \(\left(f_{2}=\right.\) \(-28.0 \mathrm{~cm}\) ). An object is placed to the left of the converging lens, and the final image produced by the two-lens combination lies \(20.7 \mathrm{~cm}\) to the left of the diverging lens. How far is the object from the converging lens?

Step-by-Step Solution

Answer

The object is approximately 47.6 cm from the converging lens.

Step 1: Understanding the System

We have two lenses: a converging lens with a focal length of \( f_1 = 24.0 \, \text{cm} \) and a diverging lens with a focal length of \( f_2 = -28.0 \, \text{cm} \). The distance between the two lenses is \( d_{12} = 56.0 \, \text{cm} \). The final image is formed \( 20.7 \, \text{cm} \) to the left of the diverging lens. We need to find the distance of the object from the converging lens.

Step 2: Calculating Final Image Distance for Second Lens

Since the final image is located \( 20.7 \, \text{cm} \) to the left of the diverging lens, we represent this image distance for the diverging lens as \( v_2 = -20.7 \, \text{cm} \) (negative because it is on the same side as the incoming light).

Step 3: Finding Image Distance from Converging Lens

Using the lens formula \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \), solve for the image distance \( v_2 \) when \( v_2 \) is the object for the diverging lens. Rearrange the formula as:\[ u_2 = \frac{f_2 v_2}{v_2 - f_2} \]Substitute \( f_2 = -28.0 \, \text{cm} \) and \( v_2 = -20.7 \, \text{cm} \):\[ u_2 = \frac{-28.0 \times (-20.7)}{-20.7 + 28.0} = 99.86 \, \text{cm} \] (from the right of the diverging lens)

Step 4: Adjusting for Lens Separation

Since the image from the first lens becomes the object for the second lens, and \( u_2 = 99.86 \text{ cm} \) is measured from the second lens, adjust for the separation between lenses:\[ v_1 = u_2 - d_{12} = 99.86 - 56.0 = 43.86 \, \text{cm}. \]This is the image distance from the first lens to the image that serves as an object for the second lens.

Step 5: Use Lens Formula to Find Object Distance

Apply the lens formula to the converging lens to find the object distance \( u_1 \) using \( f_1 = 24.0 \, \text{cm} \) and \( v_1 = 43.86 \, \text{cm} \):\[\frac{1}{f_1} = \frac{1}{u_1} + \frac{1}{v_1} \Rightarrow \frac{1}{u_1} = \frac{1}{f_1} - \frac{1}{v_1} = \frac{1}{24.0} - \frac{1}{43.86}\]Solving yields:\[ u_1 \approx 47.6 \, \text{cm} \] (object distance from the converging lens).

Key Concepts

Converging Lens

A converging lens, often known as a convex lens, is a lens that brings light rays to a focal point. When parallel rays of light pass through a converging lens, they are bent inward and converge at a point known as the focal point. This is why converging lenses are crucial in optics for focusing light.
The focal length of a converging lens is positive, indicating its ability to gather light. In our given problem, the converging lens has a focal length of 24.0 cm. This means that light rays parallel to the principal axis will converge 24.0 cm from the lens.

• The converging lens is used to form real and inverted images, which can be smaller, the same size, or larger than the object's size depending on the object's position relative to the lens.
• In practical applications, converging lenses are used in cameras, glasses, microscopes, and many other devices that need to concentrate light.

Understanding the behavior of converging lenses is essential for solving related optics problems effectively.

Diverging Lens

Diverging lenses, or concave lenses, spread out (diverge) light rays that pass through them. Unlike converging lenses, diverging lenses have a negative focal length. This means they cause parallel incoming light rays to spread out as if they originated from a point behind the lens.
In the problem, our diverging lens has a focal length of -28.0 cm. This indicates that the focal point, the point at which light rays appear to diverge, is 28.0 cm behind the lens.

• Diverging lenses are used to produce virtual images; these images cannot be projected on a screen as they appear to be located on the same side as the object.
• Such lenses are typically used in eyewear for correcting myopia (nearsightedness), in peepholes of doors, and in some camera systems to control the expansion of light rays.

The interaction between converging and diverging lenses can create a range of complex optical effects which are utilized in various optical systems.

Lens Formula

The lens formula in optics is a fundamental relationship between the object distance (u), the image distance (v), and the focal length (f) of the lens. It is given by the equation: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]This equation is crucial for determining either the image distance, the object distance, or the focal length when the other two are known.
For different types of lenses, this formula remains applicable, but interpretations will differ because the focal length for a converging lens is positive, while it is negative for a diverging lens.

• The lens formula applies to both real and virtual images, with appropriate sign conventions considered for each scenario.
• Understanding the lens formula allows you to analyze the optical power and the type of image formed, enhancing problem-solving in optical setups.

Mastering this formula is indispensable for students dealing with lens-related problems, as evidenced in our exercise.

Image Distance

Image distance, denoted as "v" in optics, is the distance from the lens to the image formed. Whether an image is formed on the same side or the opposite side of the object depends on the type of lens used and the object's initial distance.
In our exercise, realizing the image distance is helpful in tracing the path of light through the lens system. For the converging lens, the initial image distance becomes the object distance for the diverging lens.

• Using the lens formula, one can find the image distance by rearranging the relationship: \[ v = \left( \frac{f \cdot u}{u - f} \right) \]
• This allows for the determination of where and how an image will form - crucial for designing optical systems such as cameras and glasses.

Understanding how to calculate image distance for different lenses is critical for the accurate analysis and understanding of optical systems.