Problem 41 - Fundamentals of Physics

Question

A lens is made of glass having an index of refraction of 1.5. One side of the lens is flat, and the other is convex with a radius of curvature of \(20 \mathrm{~cm} .\) (a) Find the focal length of the lens. (b) If an object is placed \(40 \mathrm{~cm}\) in front of the lens, where is the image?

Step-by-Step Solution

Answer

(a) The focal length is \(-40 \text{ cm}\). (b) The image is 20 cm in front of the lens, as a virtual image.

Step 1: Understand the Lensmaker's Equation

The Lensmaker's Equation for a lens with one flat surface (where the radius of curvature is infinity) and one curved surface is \( \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \). Here, \( f \) is the focal length, \( n \) is the index of refraction, \( R_1 \) and \( R_2 \) are the radii of curvature for the two surfaces of the lens. For a plano-convex lens, \( R_1 = \infty \) (flat side) and \( R_2 = 20 \text{ cm} \) (convex side).

Step 2: Calculate the Focal Length

Substitute the known values into the Lensmaker's Equation: \( n = 1.5 \), \( R_1 = \infty \), and \( R_2 = 20 \text{ cm} \). The equation becomes \( \frac{1}{f} = (1.5-1) \left( \frac{1}{\infty} - \frac{1}{20} \right) \). Simplifying, \( \frac{1}{f} = 0.5 \times \left( -\frac{1}{20} \right) = -\frac{0.5}{20} = -\frac{1}{40} \). Thus, \( f = -40 \text{ cm} \).

Step 3: Use the Lens Formula to Locate the Image

The lens formula is \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \( v \) is the image distance and \( u \) is the object distance. Given \( u = -40 \text{ cm} \) (since the object distance is negative by sign convention for real objects), and \( f = -40 \text{ cm} \), substitute into the equation: \( \frac{1}{-40} = \frac{1}{v} - \frac{1}{-40} \). Simplify to get \( \frac{1}{v} = \frac{1}{-40} + \frac{1}{-40} = -\frac{2}{40} = -\frac{1}{20} \). Thus, \( v = -20 \text{ cm} \).

Step 4: Interpret the Image Position Result

The image distance \( v = -20 \text{ cm} \) indicates the image is formed on the same side as the object (since it is a virtual image), 20 cm in front of the lens.

Key Concepts

Focal Length

In optics, focal length is a crucial concept that determines how a lens converges or diverges light. For a lens with one flat and one curved surface, such as a plano-convex lens, the focal length is determined using the Lensmaker's Equation. In the exercise, we calculated the focal length ( f ) to be -40 cm. This tells us that the lens causes light rays to focus 40 cm away from the lens on the source side, indicating a diverging lens. Understanding focal length helps predict where an image will form, essential for designing optical systems.

Lensmaker's Equation

The Lensmaker's Equation is a key formula used in optics to calculate the focal length of a lens based on its geometry and material properties. It is given as \[\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]where \(f\) is the focal length, \(n\) is the index of refraction, and \(R_1\) and \(R_2\) are the radii of curvature of the lens surfaces. Here, one surface of the lens is flat, making \(R_1 = \infty\), and the other is convex with \(R_2 = 20\) cm. The equation simplifies dramatically with these values, helping us find the lens's behavior in focusing light. It's essential for understanding how different lenses will affect light passing through them.

Image Formation

Understanding image formation is vital in optics. When an object is placed at a certain distance from a lens, the lens will form an image at a specific location and characteristics. Image formation is determined using the lens formula:\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\]where \(v\) is the image distance, \(u\) is the object distance, and \(f\) is the focal length. For our exercise, placing the object 40 cm in front of the lens resulted in an image 20 cm in front of the lens on the same side. This means we have a virtual, upright, and reduced image, which is typical for scenarios involving diverging lenses when the object is closer than the focal length.

Index of Refraction

The index of refraction (\(n\)) is a measure of how much light slows down as it passes through a material. It is a dimensionless number that describes the bending of light as it moves from one medium to another.In this exercise, the lens is made of glass with an index of refraction of 1.5. This number tells us that light travels slower in glass than in a vacuum, leading to changes in the speed and direction of light rays.A higher index of refraction generally results in a lens having a shorter focal length, affecting how sharply it can converge or diverge light. Understanding this is fundamental in designing lenses for various optical applications, from cameras to glasses.