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A Glass Rod. Both ends of a glass rod with index of refraction 1.60 are ground and polished to convex hemispherical surfaces. The radius of curvature at the left end is 6.00 \(\mathrm{cm}\) , and the radius of curvature at the right end is 12.0 \(\mathrm{cm}\) . The length of the rod between vertices is \(40.0 \mathrm{cm} .\) The object for the surface at the left end is an arrow that lies 23.0 \(\mathrm{cm}\) to the left of the vertex of this surface. The arrow is 1.50 \(\mathrm{mm}\) tall and at right angles to the axis. (a) What constitutes the object for the surface at the right end of the rod? (b) What is the object distance for this surface? (c) Is the object for this surface real or virtual? (Hint: See Problem \(34.69 . )(\mathrm{d})\) What is the position of the final image? (e) Is the final image real or virtual? Is it erect or inverted with respect to the original object? (f) What is the height of the final image?

Step by step solution

01

Refraction at the Left Surface

Using the lens formula for refraction at a spherical surface, we have: \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \). Given, \( n_1 = 1.0 \) (air), \( n_2 = 1.6 \) (glass), \( u = -23.0 \ cm \) (left side), and \( R = 6.0 \ cm \) (convex surface), substitute the values:\[ \frac{1.6}{v} + \frac{1.0}{23} = \frac{0.6}{6} \]Solving for \( v \), which is the image distance from the left surface.

02

Calculation of Image Distance for Left Surface

Solve the equation:\[ \frac{1.6}{v} = \frac{0.6}{6} - \frac{1}{23} \]Calculate \( v \):\[ v = \frac{1.6}{\frac{0.6}{6} - \frac{1}{23}} \approx 13.33 \ cm \]This means the image is formed 13.33 cm inside the rod from the left surface.

03

Determine Object for Right Surface

The image formed by the left end serves as the object for the right end. The location within the rod is 13.33 cm from the left surface. Since the rod length is 40 cm, the object for the right end is \( 40 - 13.33 = 26.67 \ cm \) from the right surface towards the inside.

04

Object Distance for Right Surface

Since the object is inside the glass, the object distance \( u \) for the right surface will be -26.67 cm (inside towards the incident light direction) in the glass.

05

Real or Virtual Object for Right Surface

Check if the object is real or virtual. Since the object distance calculated is negative (inside the same medium), the object is virtual for the right surface.

06

Refraction at Right Surface

Apply the lens formula for the right surface:\[ \frac{1.0}{v'} - \frac{1.6}{-26.67} = \frac{1.0 - 1.6}{-12} \]Find the new image distance \( v' \).

07

Calculate Final Image Position

Solve for \( v' \):\[ \frac{1.0}{v'} = \frac{1.6}{26.67} + \frac{0.6}{12} \]Calculate \( v' \): \[ v' \approx 9.23 \ cm \]This is the final image position from the right end in air.

08

Determine Final Image Characteristics

Since \( v' \) is positive, the image is real and formed on the right side of the rod. As the image location is outside the rod, it is real. Use the magnification formula to determine if the image is erect or inverted:The initial magnification from left surface:\[ m_1 = -\frac{v}{u} = -\frac{13.33}{-23} \approx 0.58 \] (erect since positive)Final magnification for whole rod:\[ m = m_1 \times (-\frac{v'}{-u}) = 0.58 \times \frac{9.23}{26.67} \approx 0.20 \]The image is erect.

09

Calculate Final Image Height

Use the total magnification to determine final image height. The original arrow height is 1.50 mm:\[ h' = m \times h = 0.20 \times 1.5 \ mm \approx 0.30 \ mm \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Refraction

Refraction is a fundamental concept in optics that involves the bending of light as it passes from one medium to another. This bending occurs due to the change in speed of light in different materials. The degree of bending is determined by the indices of refraction of the two media involved. In this exercise, light travels from air into a glass rod, which has a higher index of refraction of 1.60. The change in medium alters the light path because glass slows down light more than air does. Understanding refraction is essential for analyzing how lenses and other optical systems redirect light to form images.

Image Formation

Image formation through lenses and curved surfaces is a key concept in optics. In the given example, each end of the glass rod acts as a lens due to its convex shape. When light passes through these surfaces, it refracts and converges or diverges to form an image. The location and nature of this image—whether it is real or virtual—are determined by how light rays converge or appear to converge. For the left surface, the image forms inside the rod due to the convex shape and is calculated to be at 13.33 cm from the lens, demonstrating how surfaces can focus light.

Lens Formula

The lens formula for a spherical surface is instrumental in solving optics problems. It is given by: \[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \] where

• \( n_1 \) and \( n_2 \) are the refractive indices of the first and second medium respectively,
• \( u \) is the object distance,
• \( v \) is the image distance, and
• \( R \) is the radius of curvature of the surface.

This equation helps us determine the position of the image formed by a lens. In the exercise, it allows us to calculate the image distances for both surfaces of the rod. By substituting the known values, we find the image distance from each surface, which is crucial in understanding how combined lenses operate.

Magnification

Magnification indicates how much an image is enlarged or reduced in size compared to the original object. It can be calculated using the formula: \[ m = -\frac{v}{u} \] for a single lens, where:

• \( m \) is the magnification,
• \( v \) is the image distance, and
• \( u \) is the object distance.

In the example, the magnification from the left surface is about 0.58, implying the image is smaller yet erect. The total magnification for the entire rod system is compounded from each surface, giving an overall magnification of 0.20. This means the final image is reduced compared to the original object size and remains upright.

Virtual and Real Images

Distinguishing between virtual and real images is important in optics. Real images are formed when light rays actually converge at a point, and can be projected onto a screen. Virtual images, on the other hand, occur when light rays appear to diverge from a point behind the lens or mirror. In this exercise, the first image formed by the left end acts as a virtual object for the right end because it is inside the rod. The outcome after the right surface is a real image since it forms outside the rod. The distinction helps in predicting the behavior of images in complex systems and understanding the lens's effect on light.