Problem 84 - University Physics with Modern Physics

Question

The radii of curvature of the surfaces of a thin converging meniscus lens are \(R_1\) = +12.0 cm and \(R_2\) = +28.0 cm. The index of refraction is 1.60. (a) Compute the position and size of the image of an object in the form of an arrow 5.00 mm tall, perpendicular to the lens axis, 45.0 cm to the left of the lens. (b) A second converging lens with the same focal length is placed 3.15 m to the right of the first. Find the position and size of the final image. Is the final image erect or inverted with respect to the original object? (c) Repeat part (b) except with the second lens 45.0 cm to the right of the first.

Step-by-Step Solution

Answer

(a) Image position from first lens is -30.0 cm; height is -3.33 mm (inverted). (b) Final image is -1.0 cm, erect. (c) Final image is -56.25 cm, inverted relative to original.

Step 1: Calculate Focal Length of the Lens

For a lens immersed in air, the focal length \( f \) is given by the lens maker's formula: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]Here, \( n = 1.60 \), \( R_1 = +12.0 \text{ cm} \), and \( R_2 = +28.0 \text{ cm} \). Substituting these values, we have:\[ \frac{1}{f} = (1.60 - 1) \left( \frac{1}{12} - \frac{1}{28} \right) \]Calculate \( \frac{1}{f} \) and thus \( f \).

Step 2: Find Image Position from First Lens

Use the lens formula to find the image position for the object:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]where \( d_o \) is the object distance (45.0 cm), and \( d_i \) is the image distance. We found \( f \) in the previous step. Substitute \( d_o \) and \( f \) to solve for \( d_i \).

Step 3: Calculate Magnification from First Lens

The magnification \( m \) is given by:\[ m = -\frac{d_i}{d_o} \]The object height \( h_o \) is 5.00 mm. The image height \( h_i \) can be calculated as:\[ h_i = m \times h_o \]Solve for \( h_i \) using the calculated magnification.

Step 4: Determine Image Position and Size with Second Lens

For the second lens placed 315 cm to the right, treat the image from the first lens as the object for this lens. Add 315 cm to the image distance of the first lens to find the new object distance for the second lens. Use the lens formula again,\[ \frac{1}{f} = \frac{1}{d_{o2}} + \frac{1}{d_{i2}} \]to find \( d_{i2} \), the image distance for the second lens. Calculate the magnification and image height again.

Step 5: Repeat for Second Lens at 45 cm

For the second lens 45 cm to the right of the first, repeat step 4 but with 45 cm added to the first image distance to get the new object distance. Use the lens formula and calculate the new image distance and height. Determine the orientation of the final image relative to the original object using the signs of magnifications from both lenses.

Key Concepts

Focal Length Calculation

The focal length of a lens is a crucial factor that determines how it focuses light. For many lenses, you can use the lens maker's formula to calculate the focal length, especially for lenses immersed in air. The formula is:
\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]Where:

• \( f \) is the focal length.
• \( n \) is the index of refraction of the lens.
• \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

By substituting the given values \( R_1 = +12.0 \text{ cm} \), \( R_2 = +28.0 \text{ cm} \), and \( n = 1.60 \), we can solve for the focal length. First, calculate each part within the parentheses before finding the reciprocal to obtain \( f \). This calculated focal length will help determine how the lens converges or diverges light rays that pass through it.

Lens Maker's Formula

The lens maker's formula is especially useful for designing lenses to achieve specific optical properties. It involves understanding how the curvature of the lens surfaces and the material's index of refraction contribute to the focal length.
For a converging lens, where both radii of curvature \( R_1 \) and \( R_2 \) are positive, the formula has a straightforward application. However, if the lens were diverging, or if one surface is convex and the other concave, the signs of the radii would be adjusted accordingly.
When using the lens maker's formula, remember:

• A positive focal length indicates a converging lens.
• A negative focal length indicates a diverging lens.
• The index of refraction affects how much the light bends; a higher index means more bending.

By correctly inputting the known values and understanding these sign conventions, you can accurately determine the lens's ability to converge or diverge light.

Image Magnification

Magnification tells us how much larger or smaller the image is compared to the object. It can also indicate whether the image is inverted or erect. The formula for magnification \( m \) is simple:
\[ m = -\frac{d_i}{d_o} \]Where:

• \( d_i \) is the image distance.
• \( d_o \) is the object distance.

The negative sign signifies that a negative magnification means the image is inverted, while a positive sign would indicate an erect image. For size calculations, the height of the image \( h_i \) can be found by multiplying the object's height \( h_o \) by the magnitude of the magnification:
\[ h_i = m \times h_o \]Understanding magnification is key in determining how images appear through lenses and in practical applications like cameras and microscopes.

Converging Lenses

Converging lenses are characterized by their ability to focus parallel light rays to a point known as the focal point. These lenses are thicker at the center than at the edges. Converging lenses are commonly used in applications where image formation is required, such as in eyeglasses, cameras, or projectors.
Key features of converging lenses include:

• A positive focal length, indicating they bring light to a focus.
• They can form real, inverted images when the object is placed outside the focal length.
• They can also form virtual, erect images when the object is within the focal point.

When working with converging lenses, using the lens formula and understanding how the focal length influences light paths are vital for accurately predicting where an image will form.

Image Orientation

The orientation of an image formed by lenses is an essential aspect of optics. This defines whether the image appears upright or flipped compared to the original object. The orientation is mainly determined by the sign of the magnification value.
Important points about image orientation:

• An erect image has a positive magnification.
• An inverted image has a negative magnification.
• A series of lenses can change the orientation multiple times, requiring careful calculation.

When dealing with multiple lenses, the overall magnification is the product of the individual magnifications. This helps determine the final orientation of an image compared to the original object, whether it has flipped, remained the same, or flipped multiple times.