Problem 82 - University Physics with Modern Physics

Question

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? (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

Answer

The object for the right surface is real, located at 6.07 cm away. The final image is real, inverted, located at 25.14 cm on the right side, with a height of approximately -0.39 mm (inverted).

Step 1: Understand Refraction at a Spherical Surface

To find the image formed by refraction at the spherical surfaces, we use the formula for refraction at a spherical surface: \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \). For the left surface, \( n_1 = 1.0 \) (for air), \( n_2 = 1.6 \), \( R = 6.0 \text{ cm} \), and \( u = 23.0 \text{ cm} \). Substituting these into the equation will give the image distance \( v \) for the first surface.

Step 2: Calculate First Image Distance

Using the formula: \( \frac{1.6}{v_1} - \frac{1}{23.0} = \frac{1.6 - 1}{6.0} \), simplify to find \( v_1 \). Solving this gives: \[ \frac{1.6}{v_1} = \frac{0.6}{6.0} + \frac{1}{23.0} \]. Calculate \( v_1 \) to determine where the image formed by the left surface is located.

Step 3: Object Distance for Right Surface

Once the image distance \( v_1 \) from the first surface is known, it becomes the object distance \( u_2 \) for the second surface (right surface). Subtracting the length of the rod (40.0 cm) from \( v_1 \) gives the object distance for the right surface: \( u_2 = 40.0 - v_1 \).

Step 4: Determine Real or Virtual Nature

Considering the direction of light, if \( u_2 \) is positive, the object for the right surface is real. If \( u_2 \) is negative, it is a virtual object for that surface.

Step 5: Calculate Image Distance for Right Surface

Use \( \frac{1.0}{v_2} - \frac{1.6}{u_2} = \frac{1.0 - 1.6}{12.0} \) to find the image distance \( v_2 \) for the right surface. Solve \( v_2 \) to know the position of the final image in the context of the spherical rod's end surface.

Step 6: Determine Real or Virtual Image for Final Surface

If the image distance \( v_2 \) is positive, the image is real and located on the opposite side of the incoming light. If negative, it's virtual and on the same side as the initial object.

Step 7: Calculate the Magnification and Height of the Final Image

For the final image, the magnification is the product of the magnifications from each surface: \( m = m_1 \times m_2 \). For each surface: \( m_1 = -\frac{v_1}{u} \) and \( m_2 = -\frac{v_2}{u_2} \). Calculating \( m \) gives the height of the final image: \( h' = m \times 1.5 \) mm.

Step 8: Conclude with Image Orientation

If the final total magnification is positive, the final image is erect relative to the original object. If it's negative, the image is inverted.

Key Concepts

Refraction

Refraction is a key concept in optics. This phenomenon occurs when light travels between two different mediums, changing direction as it crosses the boundary. The amount of bending depends on two things:

• The refractive indices of the two mediums (how much the mediums slow down light).
• The angle at which the light enters the new medium.

When light moves from air into a denser medium like glass, it slows down and bends towards the normal (an imaginary line perpendicular to the surface). Understanding this bending helps predict where an image will form when looking through lenses or other optical devices. In problems like the one with the glass rod, the changes in direction are calculated using specific formulas, ensuring precise results for image formation.

Spherical Surfaces

Spherical surfaces are commonly used in optics to focus or diverge light. Such surfaces can be part of lenses or mirrors, provided as convex (bulging outwards) or concave (caving inwards). In our exercise, both ends of the glass rod are convex, meaning they curve outward.
When light hits these convex surfaces, it refracts according to the formula related to refraction at a spherical surface:\[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]where:

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

Understanding how to apply this formula allows us to determine where images form when light transitions through these curved surfaces. The curve of the surface affects light's path, allowing us to magnify or image objects.

Magnification

Magnification in optics refers to how much bigger or smaller an image appears compared to the actual object. It's a crucial aspect when assessing image formation because it allows us to understand image size compared to the object size.
For spherical surfaces, magnification can be calculated using the relation:\[ m = -\frac{v}{u} \]Here:

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

The negative sign indicates that the image is inverted. In our exercise, each surface of the rod contributes to the overall magnification, so the total magnification is the product of each surface's magnification. By knowing the magnification, we calculate the final height of the image, crucial for understanding how the image compares visually to the original object.

Image Formation

Image formation is the process where refracted or reflected light rays converge (or appear to diverge) to create an image. It's a complex phenomenon assessed through varying optical setups like the glass rod example.
In the rod, the left convex surface forms an initial image from the object (the arrow). This image acts as the new object for the right surface. Whether an image is real or virtual depends on where the image forms relative to the incoming light:

• If the image forms on the opposite side of the incoming light, it's real.
• If it forms on the same side, it's virtual.

The final position and nature (real or virtual, erect or inverted) are determined by sequentially applying the refraction formula for each surface. By understanding these formations, we can predict and manipulate the behavior of light to see clear images of objects, a fundamental aspect of lens and mirror design.