Problem 46 - College Physics

Question

\(\cdot(\) a) You want to use a lens with a focal length of 35.0 \(\mathrm{cm}\) to produce a real image of an object, with the image twice as long as the object itself. What kind of lens do you need, and where should the object be placed? (b) Suppose you want a virtual image of the same object, with the same magnification - what kind of lens do you need, and where should the object be placed?

Step-by-Step Solution

Answer

(a) Use a convex lens; place the object 23.3 cm from the lens. (b) Use a concave lens; place the object 35 cm from the lens.

Step 1: Determine Lens Type for Real Image

To create a real image, you need a converging lens. A converging lens, often called a convex lens, can produce real images by bending the light so that it converges at a point.

Step 2: Calculate Object Distance for Real Image

For a real image with magnification \( M = -2 \) (since the image is inverted), apply the magnification formula: \( M = \frac{-d_i}{d_o} \), where \( d_i \) is the image distance and \( d_o \) is the object distance. Substitute \( M = -2 \), resulting in \( \frac{-d_i}{d_o} = -2 \), therefore \( d_i = 2d_o \).

Step 3: Use Lens Formula for Real Image

Use the lens formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Substitute \( f = 35 \text{ cm} \) and \( d_i = 2d_o \):\[ \frac{1}{35} = \frac{1}{d_o} + \frac{1}{2d_o} \].Solve this equation to find \( d_o \):Factoring gives \( \frac{1}{d_o} = \frac{3}{70} \). So, \( d_o = \frac{70}{3} \approx 23.3 \text{ cm} \).

Step 4: Determine Lens Type for Virtual Image

To create a virtual image with the same magnification \( M = 2 \), you need a diverging lens. A diverging lens, often called a concave lens, can produce virtual images where the image appears on the same side as the object.

Step 5: Use Lens Formula for Virtual Image

For virtual images, \( M = 2 \), this implies \( \frac{d_i}{d_o} = 2 \) or \( d_i = 2d_o \). However, for a virtual image, \( d_i \) is considered negative because it is on the same side as the object:Using the lens formula again: \( \frac{1}{f} = \frac{1}{d_o} - \frac{1}{2d_o} \) and \( f = 35 \). This simplifies to:\[ \frac{1}{35} = \frac{1}{d_o} - \frac{1}{2d_o} \].Solving gives \( \frac{1}{d_o} = \frac{1}{35} \). Thus, \( d_o = 35 \text{ cm} \).

Key Concepts

Converging Lens

A converging lens, which is also known as a convex lens, is essential for producing real images.
The design of a converging lens allows light rays to intersect at a focal point, which converts the light into a focused image on the opposite side of the lens.
This type of lens is characterized by its outward curving surfaces, like a football that has been slightly flattened.
Converging lenses are popularly used in applications that require magnification, such as microscopes and cameras, where real and enlarged images are beneficial.

• Create Real Images: By bending light inwards towards a point.
• Magnification: Can either enlarge or reduce the size of the image, depending on the object's distance from the lens.
• Focal Point: The point where light rays converge, this is key in determining how far an object should be from the lens to get the desired size of the image.

In the context of the given exercise, using a converging lens with a focal length of 35 cm is ideal for creating a real image that is twice the size of the object, which can be achieved by placing the object approximately 23.3 cm from the lens.

Diverging Lens

A diverging lens, known as a concave lens, is utilized when the goal is to produce virtual images.
These lenses spread light rays outwards as if they are emerging from a point behind the lens.
With a shape that curves inward, diverging lenses are employed in devices like peepholes and some glasses for vision correction, where a reduced image is required.

• Create Virtual Images: By making light rays appear to diverge from a focal point behind the lens.
• Image Formation: The image forms on the same side as the object, meaning it's virtual and cannot be captured on a screen.
• Uses: Great for applications needing a spread of light, like scatter correction in optical systems.

In the exercise scenario, a diverging lens with a focal length of 35 cm is used to produce a virtual image of the same object with the same magnification. For this, the object should be placed exactly 35 cm from the lens.

Lens Formula

The lens formula is a crucial mathematical tool used to connect the focal length of the lens, the object distance, and the image distance.
This formula is expressed as \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) represents the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance.
Understanding and applying this formula is fundamental to determining the placement of the object in relation to the lens for achieving the desired image properties.

• Equation Balance: Adjusting one variable affects the others, thus controlling the outcome of image size and position.
• Solving Problems: By rearranging and solving the formula, students can find unknown distances when given two known variables.
• Versatility: Works for both real (positive \( d_i \)) and virtual images (negative \( d_i \)).

In application, knowing the lens formula allows you to calculate the precise object placement that ensures the desired real or virtual image, whether you are using a converging or a diverging lens.