COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 24, Problem 91QAP
To determine
The index of refraction of the material.
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•48 An object is moved along the central axis of a thin lens while
the lateral magnification m is measured. Figure 34-43 gives m ver-
sus object distance p out to p, = 8.0 cm. What is the magnification
of the object when the object is 14.0 cm from the lens?
p (cm)
•8 An object is placed against the
center of a spherical mirror and
then moved 70 cm from it along the
central axis as the image distance i is
measured. Figure 34-36 gives i
versus object distance p out to P. =
40 cm. What is i for p = 70 cm?
400
-400
A diverging lens with /= -32.5 em is placed 14.0 en behind a converging
lens with -20.5 cm.
• Part A
Where will an object at infinity be focused? Determine the image distance from the second lens. Folow the sign conventions
Express your answer to two significant figures and include the appropriate units.
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Chapter 24 Solutions
COLLEGE PHYSICS
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- A camera with a 50.0 mm focal length lens is being used to photograph a person standing 8.00 m away. • How far from the lens must the film be? Provide the solution: cmarrow_forward34.40 • A diverging lens with a focal length of -48.0 cm forms a virtualimage 8.00 mm tall, 17.0 cm to the right of the lens. Determine theposition and size of the object. Is the image erect or inverted? Are theobject and image on the same side or opposite sides of the lens? Draw aprincipal-ray diagram for this situation.arrow_forward• Discussion: 1- Define refraction index? What its depending parameters? Value for air and water ? 2- Define with physical units and relationships if founded: Critical angle - Compound lens - Parallax eliminating - Lens Maker's - equation - Total Internal Reflection (T.I.R) - Snell's Law? 3- What are Real width and Virtual width? 4- What Optical Density? Is high Mass Density effect? Explain with examples? 5- Is thickness of media effected to refraction index? Why?. 6- What difference between ice and water refraction index? Which larger? Why? 7- In experiment - part one, can you find refraction index of oil? Why? Real Width r3 -r1 ng 8- Prove: r3 - r2 Virtual Widtharrow_forward
- Instructions: • For each item, show your complete solution and box your final answer with proper units. Questions: 1. What is the frequency of light that has a wavelength in water of 600 nm if the refractive index for this light is 1.33? 2. Light enters a glass plate whose index of refraction is 1.6 at an angle of incidence of 30°. What is the angle of refraction? 3. Light travelling in water (n = 1.33) strikes a glass plate at an angle of incidence of 53°; part of the beam is reflected and part is refracted. If the reflected and refracted portions makes an angle of 90° with each other, what is the index of refraction of the glass? 4. What is the critical angle for light passing from diamond to air if the index of refraction of diamond is 2.42? 5. A woman stands 2.0 m in front of a convex mirror and notices that her image height is 1/4 of her actual height. Determine the radius of curvature of the mirror.arrow_forward3. Bifocal lenses are prescribed for a patient, the components having focal length of 40cm & -300cm. • What are the new and far points of the patients eye? 4. 45 year old man can see with his left eye between 25 → 300 cm and in his right eye between 15cmo, a man has astigmatism at his left eye which requires a lens of focal length + 0.5m at axis 120°. The patient requires magnification glasses of focal length 0.25m. write the required prescription for this patientarrow_forward•• Alight ray passes through a prism with an apex angle of a, as shown in Figure 31-62. The ray and the bisector of the apex angle intersect at right angles. Show that the angle of deviation & is related to the apex angle and the index of refraction of the prism material by sin {(a + 8)] = n sin(ta). 78arrow_forward
- • Part A Find the foca length of the glass lens in the figure (Egure 1) Take R22 cm and Ry= 40 cm Express your answer to two significant figures and include the appropriate units. 3 Figure 135.27 cm Submit Prei A Bequest.Aner X Incorrect: Try Again: 3 attempts remaining Proe Feedtadkarrow_forward• Analyze the different types of mirrors and solve problems involving mirror equation. Differentiate the two types of lenses and solve problems involving the lens equation. OBJECTIVE Identify the parts of a concave and convex lenses and mirrors. Determine the location, orientation, size and type of image produced by a concave and convex lenses and mirrors. Solve problems involving mirrors and lenses. WHAT'S IN Light is one of the necessities of life on Earth. • Light is essential in the process of photosynthesis by plants. Light is also how reception and transmission of information about the things around us become possible. WHAT'S NEW Activity 1. Pinpoint Me? Directions: Identify the words described by the statements below and write your answer on the space before each number. 1. The law which states that the angle of incidence is equal to the angle of reflection. 2. The term used to describe the scattered reflection of light. 3. The center of the sphere to which the mirror is a part.…arrow_forward•• A slab of glass that has an index of refraction of 1.50 is submerged in water that has an index of refraction of 1.33. Light in the water is incident on the glass. Find the angle of refraction if the angle of incidence is (a) 60°, (b) 45°, and (c) 30°. SSM 31arrow_forward
- Calculate the angle of refraction at the air/core interface. • nair = 1, ncore = 1.47, ncladding = 1.45, incident = 30oarrow_forward400 •44 An object is placed against the center of a thin lens and then moved away from it along the central axis as the image distance i is measured. Figure 34-41 gives i versus object distance p out to p, = 60 cm. What is the image distance when p= 100 -400 p (cm) Figure 34-41 Problem 44. cm? (un)!arrow_forward•• ENGINEERING APPLICATION An optical fiber allows rays of light to propagate long distances by using total internal reflec- tion. Optical fibers are used extensively in medicine and in digital communications. As shown in Figure 31-58 the fiber consists of a core material that has an index of refraction n, and radius b sur- rounded by a cladding material that has an index of refraction n3 < n2. The numerical aperture of the fiber is defined as sine, where 0, is the angle of incidence of a ray of light that impinges on the center of the end of the fiber and then reflects off the core- cladding interface just at the critical angle. Using the figure as a guide, show that the numerical aperture is given by sine, = Vn3 - n? assuming the ray is initially in air. Hint: Use of the Pythagorean theorem may be required. 44arrow_forward
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Laws of Refraction of Light | Don't Memorise; Author: Don't Memorise;https://www.youtube.com/watch?v=4l2thi5_84o;License: Standard YouTube License, CC-BY