Problem 22 - University Physics with Modern Physics

Question

The left end of a long glass rod 8.00 \(\mathrm{cm}\) in diameter, with an index of refraction of \(1.60,\) is ground and polished to a convex hemispherical surface with a radius of 4.00 \(\mathrm{cm} .\) An object in the form of an arrow 1.50 \(\mathrm{mm}\) tall, at right angles to the axis of the rod, is located on the axis 24.0 \(\mathrm{cm}\) to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?

Step-by-Step Solution

Answer

The image is 8.35 cm to the right, 0.522 mm tall, and erect.

Step 1: Identify the Known Values

We are given:- Diameter of the rod: 8.00 cm- Index of refraction, \( n = 1.60 \)- Radius of the convex surface, \( R = 4.00 \) cm- Object distance from the vertex, \( s = 24.0 \) cm- Height of the object, \( h_{o} = 1.50 \) mmWe need to find: the image position (\( s' \)), the height of the image (\( h' \)), and whether the image is erect or inverted.

Step 2: Refraction Formula for Spherical Surfaces

Use the refraction formula for spherical surfaces: \[ \frac{n_1}{s} + \frac{n_2}{s'} = \frac{n_2 - n_1}{R} \]where:- \( n_1 = 1 \) (air)- \( n_2 = 1.60 \) (glass rod)- \( s = -24.0 \) cm (negative since object is on the same side as the incoming light)- \( R = 4.00 \) cmSubstitute the values into the equation to solve for \( s' \).

Step 3: Substitute and Solve for Image Position

Substitute the known values into the formula:\[ \frac{1}{-24.0} + \frac{1.60}{s'} = \frac{1.60 - 1}{4.00} \]This simplifies to:\[ \frac{1}{-24.0} + \frac{1.60}{s'} = \frac{0.60}{4.00} \]Further simplify to:\[ -\frac{1}{24} + \frac{1.60}{s'} = 0.15 \]Clear the fraction and solve for \( s' \).

Step 4: Calculate the Value of s'

Continuing from the previous equation:\[ \frac{1.60}{s'} = 0.15 + \frac{1}{24} \]\[ 0.15 = \frac{15}{100} + \frac{1}{24} \]Find a common denominator and solve:\[ \frac{15}{100} = \frac{18}{120}, \quad \frac{1}{24} = \frac{5}{120} \]\[ 0.15 + \frac{1}{24} = \frac{18 + 5}{120} = \frac{23}{120} \]Now, \( \frac{1.60}{s'} = \frac{23}{120} \), so solve for \( s' \):\[ s' = \frac{1.60 \times 120}{23} = \frac{192}{23} \approx 8.35 \text{ cm} \]

Step 5: Determine Image Characteristics

Use the magnification formula: \[ m = \frac{h'}{h_o} = -\frac{s'}{s} \]Substitute the values:\[ m = -\frac{8.35}{-24} \approx 0.348 \]Since \( m > 0 \), the image is erect.The image height is:\[ h' = m \times h_o \approx 0.348 \times 1.50 \text{ mm} \approx 0.522 \text{ mm} \]

Key Concepts

Refraction

In optics, refraction is the bending of a light ray when it passes through the boundary between two different media. This occurs because light travels at different speeds in different media. For instance, light slows down as it passes from air into glass, due to glass having a higher optical density. This change in speed causes the light ray to bend, or refract. Refraction is governed by Snell's Law, which states:\[ n_1 \sin \theta_1 = n_2 \sin \theta_2\]where \( n_1 \) and \( n_2 \) are the indices of refraction for the original and second medium, respectively. Even though this equation is primarily used for flat surfaces, the fundamental idea of differing speeds and bending remains applicable to spherical surfaces as well.

Spherical Surfaces

Spherical surfaces, such as the convex surface of a glass rod, can focus or diverge light rays. These surfaces are portions of spheres and their curvature affects how light refracts through them. The key element in understanding refraction through spherical surfaces is the radius of curvature, \( R \). The refraction formula for spherical surfaces is:\[ \frac{n_1}{s} + \frac{n_2}{s'} = \frac{n_2 - n_1}{R}\]In this equation:

• \( n_1 \) is the refractive index of the medium the light comes from (like air).
• \( n_2 \) is the refractive index of the medium the light enters (like glass).
• \( s \) is the distance from the object to the surface.
• \( s' \) is the image distance from the surface to the image.
• \( R \) is the radius of the spherical surface.

By understanding these relationships, we can determine where an image will form relative to the surface.

Magnification

Magnification describes how much larger or smaller an image is compared to the actual object. This is crucial in optics because refraction through spherical surfaces can change the perceived size of an object. Magnification is given by the formula:\[ m = \frac{h'}{h_o} = -\frac{s'}{s}\]Where:

• \( m \) is the magnification factor.
• \( h' \) is the height of the image.
• \( h_o \) is the height of the object.
• \( s' \) is the image distance.
• \( s \) is the object distance.

A positive magnification means the image is erect, while a negative magnification indicates an inverted image. In this case, since the magnification is positive, the image is erect but smaller than the object.

Image Formation

Image formation by lenses or spherical surfaces occurs as light rays originating from an object converge or diverge to form an image. In the case of spherical surfaces, the image position and height are determined by how the light refracts. For our exercises, the use of the refraction formula gives us the image distance \( s' \), while magnification tells us the image height \( h' \).

By substituting the known values—such as refractive indices, object distance, and surface curvature—into the formula, we pinpoint where and how the image forms. This includes understanding whether the image is real or virtual, and erect or inverted. Real images occur when light converges at a point, while virtual images form when light diverges and our eyes backward-extend the rays to perceive an image.