Ch 34: Ray Optics

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 34: Ray Optics

Problem 80b

Chapter 34, Problem 80b

Consider a lens having index of refraction n₂ and surfaces with radii R₁ and R₂. The lens is immersed in a fluid that has index of refraction n₁. A symmetric converging glass lens (i.e., two equally curved surfaces) has two surfaces with radii of 40 cm. Find the focal length of this lens in air and the focal length of this lens in water.

Verified step by step guidance

Step 1: Understand the lens maker's formula, which is used to calculate the focal length of a lens. The formula is:

\frac{1}{f} = (n - 1) [\frac{1}{R} - \frac{1}{R}]

, where

n

is the index of refraction of the lens material relative to the surrounding medium, and

R 1

and

R 2

are the radii of curvature of the two surfaces of the lens.

Step 2: For the lens in air, calculate the relative index of refraction

n

n = \frac{n}{2}

, where

n 2

is the index of refraction of the lens material and

n 1

is the index of refraction of air (approximately 1). Substitute the given values into the formula.

Step 3: For the lens in water, calculate the relative index of refraction

n

n = \frac{n}{2}

, where

n 1

is the index of refraction of water (approximately 1.33). Substitute the given values into the formula.

Step 4: Substitute the radii of curvature

R 1

and

R 2

into the lens maker's formula. Since the lens is symmetric,

R 1

and

R 2

are equal in magnitude but opposite in sign. Use

R 1 = 40

cm and

R 2 = - 40

cm.

Step 5: Solve for the focal length

f

in both air and water using the respective relative indices of refraction and the substituted values for

R 1

and

R 2

. Ensure the units are consistent throughout the calculation.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Lens Maker's Formula

The Lens Maker's Formula relates the focal length of a lens to the indices of refraction of the lens material and the surrounding medium, as well as the radii of curvature of its surfaces. It is expressed as 1/f = (n₂ - n₁) * (1/R₁ - 1/R₂), where f is the focal length, n₂ is the lens's refractive index, n₁ is the surrounding medium's refractive index, and R₁ and R₂ are the radii of curvature. This formula is essential for calculating how a lens focuses light in different media.

Index of Refraction

The index of refraction (n) is a dimensionless number that describes how light propagates through a medium. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. A higher index indicates that light travels slower in that medium, affecting how light bends when entering or exiting the lens. Understanding the indices of refraction for both the lens and the surrounding fluid is crucial for accurate focal length calculations.

Focal Length

The focal length of a lens is the distance from the lens at which parallel rays of light converge (for converging lenses) or appear to diverge (for diverging lenses). It is a key characteristic that determines the lens's ability to focus light and is influenced by the lens's shape and the refractive indices of the lens material and surrounding medium. The focal length changes when the lens is placed in different media, such as air or water, which is important for solving the given problem.

Recommended video:

Guided course

10:54

Spinning on a string of variable length

Related Practice

Textbook Question

The mirror in FIGURE CP34.79 is covered with a piece of glass whose thickness at the center equals the mirror's radius of curvature. A point source of light is outside the glass. How far from the mirror is the image of this source?

views

Textbook Question

CALC FIGURE CP34.81 shows a light ray that travels from point A to point B. The ray crosses the boundary at position x, making angles θ₁ and θ₂ in the two media. Suppose that you did not know Snell's law. You've proven that Snell's law is equivalent to the statement that 'light traveling between two points follows the path that requires the shortest time.' This interesting way of thinking about refraction is called Fermat's principle. Write an expression for the time t it takes the light ray to travel from A to B. Your expression should be in terms of the distances a, b, and w; the variable x; and the indices of refraction n₁ and n₂.

views

Textbook Question

Some electro-optic materials can change their index of refraction in response to an applied voltage. Suppose a plano-convex lens (flat on one side, a 15.0 cm radius of curvature on the other), made from a material whose normal index of refraction is 1.500, is creating an image of an object that is 50.0 cm from the lens. By how much would the index of refraction need to be increased to move the image 5.0 cm closer to the lens?

views

Textbook Question

A fortune teller's 'crystal ball' (actually just glass) is 10 cm in diameter. Her secret ring is placed 6.0 cm from the edge of the ball. The crystal ball is removed and a thin lens is placed where the center of the ball had been. If the image is still in the same position, what is the focal length of the lens?

views

Textbook Question

A 2.0-cm-tall object is placed in front of a mirror. A 1.0-cm-tall upright image is formed behind the mirror, 150 cm from the object. What is the focal length of the mirror?

views

Textbook Question

A fortune teller's 'crystal ball' (actually just glass) is 10 cm in diameter. Her secret ring is placed 6.0 cm from the edge of the ball. An image of the ring appears on the opposite side of the crystal ball. How far is the image from the center of the ball?

views