Textbook Question
A lensmaker wants to make a magnifying glass from glass that has an index of refraction n = 1.55 and a focal length of 20.0 cm. If the two surfaces of the lens are to have equal radii, what should that radius be?
1989
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Ch 34: Geometric Optics
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610
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Table of contents
- Ch 01: Units, Physical Quantities & Vectors41
- Ch 02: Motion Along a Straight Line67
- Ch 03: Motion in Two or Three Dimensions47
- Ch 04: Newton's Laws of Motion23
- Ch 05: Applying Newton's Laws59
- Ch 06: Work & Kinetic Energy14
- Ch 07: Potential Energy & Conservation8
- Ch 08: Momentum, Impulse, and Collisions18
- Ch 09: Rotation of Rigid Bodies42
- Ch 10: Dynamics of Rotational Motion48
- Ch 11: Equilibrium & Elasticity27
- Ch 12: Fluid Mechanics31
- Ch 13: Gravitation35
- Ch 14: Periodic Motion32
- Ch 15: Mechanical Waves25
- Ch 16: Sound & Hearing32
- Ch 17: Temperature and Heat36
- Ch 18: Thermal Properties of Matter38
- Ch 19: The First Law of Thermodynamics33
- Ch 20: The Second Law of Thermodynamics22
- Ch 21: Electric Charge and Electric Field36
- Ch 22: Gauss' Law24
- Ch 23: Electric Potential31
- Ch 24: Capacitance and Dielectrics18
- Ch 25: Current, Resistance, and EMF26
- Ch 26: Direct-Current Circuits30
- Ch 27: Magnetic Field and Magnetic Forces18
- Ch 28: Sources of Magnetic Field10
- Ch 29: Electromagnetic Induction28
- Ch 30: Inductance17
- Ch 31: Alternating Current21
- Ch 32: Electromagnetic Waves1
- Ch 33: The Nature and Propagation of Light20
- Ch 34: Geometric Optics34
- Ch 35: Interference19
- Ch 36: Diffraction25
- Ch 37: Special Relativity20
- Ch 38: Photons: Light Waves Behaving as Particles15
- Ch 39: Particles Behaving as Waves26
- Ch 40: Quantum Mechanics I: Wave Functions21
- Ch 41: Quantum Mechanics II: Atomic Structure25
- Ch 42: Molecules and Condensed Matter18
- Ch 43: Nuclear Physics16
- Ch 44: Particle Physics and Cosmology23
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 34, Problem 37b
For each thin lens shown in Fig. E34.37, calculate the location of the image of an object that is 18.0 cm to the left of the lens. The lens material has a refractive index of 1.50, and the radii of curvature shown are only the magnitudes.
Verified step by step guidance1
Identify the lens type and use the lens maker's formula to calculate the focal length of the lens. The lens maker's formula is:
Substitute the given values into the lens maker's formula. For this problem, n = 1.50, and the radii of curvature (R1 and R2) are provided in the figure. Be sure to assign the correct signs to R1 and R2 based on the lens's geometry (convex surfaces are positive, concave surfaces are negative).
Once the focal length (f) is determined, use the thin lens equation to find the image location. The thin lens equation is:
Rearrange the thin lens equation to solve for the image distance (d_i):
Simplify the equation to find the value of , which represents the location of the image. Ensure the sign of d_i is interpreted correctly to determine whether the image is real or virtual and on which side of the lens it is located.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Thin Lens Formula
The thin lens formula relates the object distance (do), image distance (di), and focal length (f) of a lens. It is expressed as 1/f = 1/do + 1/di. This formula is essential for determining the position of the image formed by the lens based on the position of the object and the lens's focal length.
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Thin Lens Equation
Refractive Index
The refractive index (n) of a material indicates how much light slows down when passing through it compared to a vacuum. For a lens, the refractive index affects the focal length, which can be calculated using the lensmaker's equation. In this case, the refractive index of 1.50 suggests that light travels slower in the lens material than in air.
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Guided course
03:46Index of Refraction
Lensmaker's Equation
The lensmaker's equation provides a relationship between the focal length of a lens, its radii of curvature, and its refractive index. It is given by 1/f = (n - 1)(1/R1 - 1/R2), where R1 and R2 are the radii of curvature of the lens surfaces. This equation is crucial for calculating the focal length needed to apply the thin lens formula.
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05:38Lens Maker Equation
Related Practice
Textbook Question
A 1.20 cm tall object is 50.0 cm to the left of a converging lens of focal length 40.0 cm. A second converging lens, this one having a focal length of 60.0 cm, is located 300.0 cm to the right of the first lens along the same optic axis. Find the location and height of the image (call it I1) formed by the lens with a focal length of 40.0 cm.
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Textbook Question
An object is 16.0 cm to the left of a lens. The lens forms an image 36.0 cm to the right of the lens. Draw a principal-ray diagram.
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Textbook Question
A converging lens with a focal length of 70.0 cm forms an image of a 3.20 cm tall real object that is to the left of the lens. The image is 4.50 cm tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?
2013
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Textbook Question
An object is 16.0 cm to the left of a lens. The lens forms an image 36.0 cm to the right of the lens. If the object is 8.00 mm tall, how tall is the image? Is it erect or inverted?
2170
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Textbook Question
A converging lens with a focal length of 9.00 cm forms an image of a 4.00 mm tall real object that is to the left of the lens. The image is 1.30 cm tall and erect. Where are the object and image located? Is the image real or virtual?
1912
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