Problem 88 - University Physics with Modern Physics

Question

A transparent rod 50.0 \(\mathrm{cm}\) long and with a refractive index of 1.60 is cut flat at the right end and rounded to a hemispherical surface with a 15.0 -cm radius at the left end. An object is placed on the axis of the rod 12.0 \(\mathrm{cm}\) to the left of the vertex of the hemispherical end. (a) What is the position of the final image? (b) What is its magnification?

Step-by-Step Solution

Answer

(a) The final image position is 39.3 cm inside the rod from the flat end. (b) The magnification is 0.87.

Step 1: Calculate the position of the image formed by the hemispherical surface

We'll use the lens maker's formula for the hemispherical surface to find the image position, \(i_1\): \[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]Given, \(n_2 = 1.60\) (refractive index of the rod), \(n_1 = 1.00\) (refractive index of air), \(u = -12.0\, \mathrm{cm}\) (object position), and \(R = 15.0\, \mathrm{cm}\).Substitute the given values:\[ \frac{1.60}{v} - \frac{1.00}{-12.0} = \frac{1.60 - 1.00}{15.0} \]\[ \frac{1.60}{v} + \frac{1.00}{12.0} = \frac{0.60}{15.0} \]Solve for \(v\) to find the image position \(i_1\).

Step 2: Calculate the position of the image formed by the flat surface

The image from the hemispherical surface acts as the object for the flat surface, i.e., \(u_2 = 50.0 - v_1\) if \(v_1\) is to the right of the hemispherical surface.Since the flat surface is plane, apply:\[ n_2 \cdot u_2 = n_3 \cdot i_2 \]Where \(n_2 = 1.60\) and \(n_3 = 1.00\).This simplifies to:\[ 1.60 \cdot (50.0 - v_1) = 1.00 \cdot i_2 \]Solve for \(i_2\) to find the final image position.

Step 3: Calculate the total magnification

Magnification (\(M\)) is the product of the magnification at each surface.For the hemispherical surface:\[ M_1 = \frac{n_1 \cdot v_1}{n_2 \cdot u} \]For the flat surface, magnification is determined by using \(M_2 = 1\) because the flat surface doesn't change size.The total magnification is:\[ M = M_1 \cdot M_2 \]Substitute \(M_1\) and \(M_2\) to calculate \(M\).

Key Concepts

Refractive Index

The refractive index is a crucial concept in understanding how light behaves as it moves from one medium to another. It is denoted by the symbol "n" and defines the ratio of the speed of light in vacuum to the speed of light in the medium. The formula to determine it is:

• Definition: \[ n = \frac{c}{v} \]Where:
  c is the speed of light in vacuum (approximately \( 3 \times 10^8 \, \text{m/s} \))
  v is the speed of light in the medium.

A higher refractive index indicates that light travels slower in the medium compared to a vacuum. In the exercise, the refractive index of the rod is 1.60, which means light slows down significantly as it enters the rod from the air. Understanding the refractive index helps us predict how much a material will deviate light rays passing through it, which is vital for the calculation of image positions in lenses and other optical devices.

Lens Maker's Formula

The lens maker's formula is essential in geometric optics to determine the focal length of a lens from its radius of curvature and refractive index. Although traditionally used for lenses, it also provides insight into a hemispherical surface's behavior, as in the given exercise.

• Formula:\[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]
• Where:
  n_2 is the refractive index of the lens material,
  n_1 is the refractive index of air,
  R is the radius of curvature,
  u is the object distance from the surface,
  v is the image distance from the surface.

In our case, the hemispherical lens with a curvature radius of 15 cm focuses light to form an image inside the rod. Using the lens maker's formula allows us to find where the image appears.

Magnification

Magnification provides insight into how much larger or smaller an image is relative to the original object. In optics, it is a pivotal measure, especially when determining the final appearance of an image formed by a lens or lens system.

• Definition: \[ M = \frac{h'}{h} = \frac{-v}{u} \]Where:
  h' is the height of the image,
  h is the height of the object,
  v is the image distance,
  u is the object distance.

For a system with multiple elements, like our rod with both a hemispherical and flat surface, the total magnification is the product of each stage's magnification. In our specific case, since the flat surface does not cause magnification (\( M_2 = 1 \)), the total magnification is primarily influenced by the hemispherical part.

Hemispherical Lens

A hemispherical lens offers unique properties in optics, thanks to its half-sphere shape, which affects light differently than a standard lens. In geometric optics, it's used to bend light to either converge or diverge based on its curvature.

• Properties:
  • Its curvature focuses light into a single point.
  • Radius determines the lens's power to converge or diverge light rays.

In the exercise, an object placed outside the hemispherical lens results in the light being bent and focused inside the rod, calculating the image's position and characteristics. This understanding helps solve complex optics problems by defining where the image will ultimately form, using its experimented capabilities in practical applications like lenses in devices.

Optics Problem Solving

Solving optics problems involves a systematic approach to finding how light interacts with surfaces and lenses. Key steps aid in navigating through complex optics situations.

• Steps for Problem Solving:
  • Identify the types and roles of each optical surface involved.
  • Use relevant formulas, like the lens maker’s formula, to ascertain basic parameters.
  • Calculate distances and magnification for image formation at each surface.
  • Combine these calculations for composite systems to find the final image properties.

By following these methods, we can confidently determine the properties of interest, such as image position and magnification, ensuring a thorough understanding and accurate solution. For instance, in the exercise, applying these techniques allowed for calculating the image's precise location and size using the provided lens and refractive indices.