Problem 22 - University Physics with Modern Physics

Question

The left end of a long glass rod 6.00 cm in diameter has a convex hemispherical surface 3.00 cm in radius. The refractive index of the glass is 1.60. Determine the position of the image if an object is placed in air on the axis of the rod at the following distances to the left of the vertex of the curved end: (a) infinitely far, (b) 12.0 cm; (c) 2.00 cm.

Step-by-Step Solution

Answer

(a) 8.00 cm inside glass; (b) 5.65 cm inside glass; (c) 2.29 cm inside glass.

Step 1: Understand the Problem

We need to find the position of the image for an object placed at different distances from a convex hemispherical surface of a glass rod. The refractive index of the glass is given, and the curvature of the surface is specified by its radius.

Step 2: Use Lensmaker's Equation for Spherical Surfaces

The formula for refraction at a spherical surface is given by \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \), where:- \( n_1 = 1 \) (refractive index of air),- \( n_2 = 1.6 \) (refractive index of the glass),- \( R = 3.00 \) cm (radius of curvature),- \( u \) is the object distance (positive for real object in air),- \( v \) is the image distance.

Step 3: Solve for Part (a) - Object at Infinity

For an object at infinity (\( u = \infty \)), the equation simplifies to finding the focal length (\( f \)):\[ \frac{n_2}{f} = \frac{n_2 - n_1}{R} \]Calculate:\[ \frac{1.6}{f} = \frac{1.6 - 1.0}{3.00} \]\[ f = \frac{1.6 \times 3.00}{0.6} = 8.00 \text{ cm} \]Thus, the image is formed at the focal point, 8.00 cm inside the glass.

Step 4: Solve for Part (b) - Object at 12.0 cm

Substitute \( u = 12.0 \) cm in the refraction formula:\[ \frac{1.6}{v} - \frac{1}{12.0} = \frac{0.6}{3.00} \]Calculate:\[ \frac{1.6}{v} = \frac{1}{12} + \frac{0.6}{3.0} \]\[ \frac{1.6}{v} = \frac{1}{12} + 0.2 \]\[ \frac{1.6}{v} = \frac{1 + 2.4}{12} = \frac{3.4}{12} \]\[ v = \frac{1.6 \times 12}{3.4} \approx 5.65 \text{ cm} \]Thus, the image is formed 5.65 cm inside the glass.

Step 5: Solve for Part (c) - Object at 2.00 cm

Substitute \( u = 2.00 \) cm in the refraction formula:\[ \frac{1.6}{v} - \frac{1}{2.00} = \frac{0.6}{3.00} \]Calculate:\[ \frac{1.6}{v} = \frac{1}{2.00} + 0.2 \]\[ \frac{1.6}{v} = 0.5 + 0.2 \]\[ \frac{1.6}{v} = 0.7 \]\[ v = \frac{1.6}{0.7} \approx 2.29 \text{ cm} \]Thus, the image is formed 2.29 cm inside the glass.

Key Concepts

Refractive Index

The refractive index is a key concept in optical physics. It describes how light bends when it enters a different medium. For example, when light travels from air, with a refractive index of 1.0, into glass with a refractive index of 1.6, the light slows down and bends towards the normal. This bending is due to the change in speed of light in different media.

• A higher refractive index in a material means light travels slower through it.
• It affects how lenses and surfaces bend light to form images.

Understanding this concept is critical for predicting how lenses will focus light, which is essential in optical devices like glasses, cameras, and telescopes.

Spherical Surfaces

Spherical surfaces are curved surfaces shaped like a segment of a sphere. They have specific properties that affect how they refract light. In optics, these surfaces include lenses and mirrors that are used to manipulate light in various ways.

• A convex spherical surface bulges outward like the surface of a ball.
• Convex surfaces are used to converge light, such as in magnifying glasses or focusing lenses.

The radius of the spherical surface is the key measurement, indicating the curvature's extent. Smaller radii result in more pronounced curves, significantly affecting how light is bent.

Optical Image Formation

Image formation in optics involves the manipulation of light waves to produce a replica of the viewed object. When light rays pass through or reflect off surfaces, they are bent and shifted to create an image.
To determine where an image forms, you can use formulas like the Lensmaker's Equation. This equation considers variables such as:

• The refractive index of the media involved (e.g., air and glass in our problem).
• The radii of curvature of the providing surfaces.
• The positions (distances) of the object and image relative to the lens or surface.

Depending on these factors, images can be formed real, inverted, smaller, or larger than the actual object. Understanding this helps in designing lenses for eyeglasses, cameras, and many scientific instruments.