Problem 90 - University Physics with Modern Physics

Question

The radii of curvature of the surfaces of a thin converging meniscus lens are \(R_{1}=+12.0 \mathrm{cm}\) and \(R_{2}=+28.0 \mathrm{cm} .\) 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 \(\mathrm{mm}\) tall, perpendicular to the lens axis, 45.0 \(\mathrm{cm}\) to the left of the lens. (b) A second converging lens with the same focal length is placed 3.15 \(\mathrm{m}\) 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 \(\mathrm{cm}\) to the right of the first.

Step-by-Step Solution

Answer

The final images are: (b) at 20.9 cm inverted, 0.265 mm, (c) virtual at -43.3 cm, erect, 11.9 mm.

Step 1: Determine the Lens Maker's Formula

To find the focal length \( f \) of the lens, use the lens maker's formula:\[ \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]with \( n = 1.60 \), \( R_1 = +12.0\, \text{cm} \), and \( R_2 = +28.0\, \text{cm} \). Substituting these values, we get:\[ \frac{1}{f} = (1.60-1)\left( \frac{1}{12.0} - \frac{1}{28.0} \right) \]Calculating this gives \( f \approx 19.2\, \text{cm} \).

Step 2: Use the Lens Formula to Find Image Position

With the object distance \( d_o = 45.0\, \text{cm} \) (to the left, hence \( d_o = -45.0 \)), use the lens formula:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Substitute \( f = 19.2\, \text{cm} \) and \( d_o = -45.0\, \text{cm} \):\[ \frac{1}{19.2} = \frac{1}{-45.0} + \frac{1}{d_i} \]Solving for \( d_i \) gives \( d_i \approx 32.0\, \text{cm} \).

Step 3: Calculate Image Size Using Magnification

The magnification \( m \) is given by:\[ m = -\frac{d_i}{d_o} \]and the image height \( h_i \) is related to the object height \( h_o = 5.00\, \text{mm} \) by:\[ h_i = m \times h_o \]Substitute \( d_i = 32.0\, \text{cm} \), \( d_o = -45.0\, \text{cm} \), and \( h_o = 5.00\, \text{mm} \):\[ m = -\frac{32.0}{-45.0} \approx 0.711 \]\[ h_i = 0.711 \times 5.00 \approx 3.56\, \text{mm} \]

Step 4: Determine Final Image Position for Second Lens at 3.15 m

The image from the first lens acts as the object for the second lens, placed \( 3.15\, \text{m} = 315\, \text{cm} \) to the right of the first. The object distance for the second lens \( d_{o2} = 315\, \text{cm} - 32.0\, \text{cm} = 283\, \text{cm} \). Use the lens formula for the second lens with \( f = 19.2\, \text{cm} \):\[ \frac{1}{19.2} = \frac{1}{283} + \frac{1}{d_{i2}} \]Solving gives \( d_{i2} \approx 20.9\, \text{cm} \) to the right of the second lens.

Step 5: Calculate Final Image Size and Orientation

The magnification for the second lens \( m_2 \) is:\[ m_2 = -\frac{d_{i2}}{d_{o2}} = -\frac{20.9}{283} \approx -0.074 \]Combine with the magnification of the first lens \( m = 0.711 \):\[ m_{total} = m \times m_2 = 0.711 \times (-0.074) \approx -0.053 \]The final image height \( h_{i2} \) is:\[ h_{i2} = m_{total} \times h_o \approx -0.053 \times 5.00 \approx -0.265\, \text{mm} \]The final image is inverted.

Step 6: Repeat for Second Lens at 45 cm

For the second lens at \( 45.0\, \text{cm} \) to the right of the first, \( d_{o2} = 45.0\, \text{cm} - 32.0\, \text{cm} = 13.0\, \text{cm} \). Using the lens formula:\[ \frac{1}{19.2} = \frac{1}{13.0} + \frac{1}{d_{i2}} \]Solving for \( d_{i2} \) gives \( d_{i2} \approx -43.3\, \text{cm} \), meaning the image is virtual and located on the same side as the incoming light.

Step 7: Calculate Final Image Size and Orientation for Lens at 45 cm

Magnification for the second position:\[ m_2 = -\frac{d_{i2}}{d_{o2}} = -\frac{-43.3}{13.0} \approx 3.33 \]Total magnification:\[ m_{total} = 0.711 \times 3.33 \approx 2.37 \]The final image height \( h_{i2} \) is:\[ h_{i2} = m_{total} \times h_o \approx 2.37 \times 5.00 \approx 11.9\, \text{mm} \]The final image is erect.

Key Concepts

Lens Maker's Formula

The lens maker's formula is instrumental in calculating the focal length of a thin lens. This formula connects the physical attributes of the lens—specifically, the radii of curvature of its two surfaces—and the refractive index of the material the lens is made from. By knowing these properties, you can determine how strongly the lens will converge or diverge light.

The lens maker's formula is given by:
\[ \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where:

• \( f \) is the focal length of the lens.
• \( n \) is the refractive index of the lens material.
• \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces, with signs determined by the lens shape.

For the mentioned exercise, substituting \( n = 1.60 \), \( R_1 = +12.0 \text{ cm} \), and \( R_2 = +28.0 \text{ cm} \) into the formula gives a focal length of approximately \( f \approx 19.2 \text{ cm} \). This means the lens focuses parallel light rays 19.2 cm from its center.

Image Magnification

Image magnification refers to how much larger or smaller an image will appear compared to the original object. It's a critical concept in optics because it helps in understanding how lenses affect the size of the images formed.

Magnification \( m \) can be calculated using the formula:
\[ m = -\frac{d_i}{d_o} \]
where:

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

The negative sign indicates that a positive magnification means an inverted image, while a negative magnification means an upright image.

For an object placed 45.0 cm from the lens, using \( d_i \approx 32.0 \text{ cm} \), the magnification is \( m \approx 0.711 \). This results in an image that is smaller than the object, with a height of approximately 3.56 mm.

Thin Lens Equation

The thin lens equation provides a simple relation to find the position of the image formed by a lens. It's widely used because it connects the object distance, image distance, and the focal length of the lens effectively through a simple relation:

\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]
This compact formula is essential in lens optics for determining where an image will form given an object's position and the lens's focal length.

• \( f \) is the focal length of the lens.
• \( d_o \) is the object distance from the lens. It is conventionally negative if the object is on the same side as the incoming light (real object).
• \( d_i \) is the distance from the lens to the image formed. It can be positive for a real image or negative for a virtual image.

Using the object distance of \( -45.0 \text{ cm} \) and the focal length of \( 19.2 \text{ cm} \), we calculated that the image is formed at \( d_i \approx 32.0 \text{ cm} \) to the opposite side of the lens.