Concept explainers
Figure P38.76 shows an object placed a distance do1 from one of two converging lenses separated by s = 1.00 m. The first lens has focal length f1 = 22.0 cm, and the second lens has focal length f2 = 45.0 cm. An image is formed by light passing through both lenses at a distance di2 = 15.0 cm to the left of the second lens. a. What is the value of do1 that will result in this image position? b. Is the final image formed by the two lenses real or virtual? c. What is the magnification of the final image? d. Is the final image upright or inverted?
Figure P38.76
(a)
The value of
Answer to Problem 76PQ
The value of
Explanation of Solution
Write the thin lens equation for first lens.
Here,
Rearrange the above equation to find
Similarly,
The image of first lens acts as object for second lens. Hence, object distance for second lens is,
Conclusion:
Substitute
From equation (III), we get,
Substitute
Substitute
Thus, the value of
(b)
Whether the image formed by the two lenses is real or virtual.
Answer to Problem 76PQ
The image formed by the two lenses is Virtual.
Explanation of Solution
Since, the image is formed on the left side of the second lens, the image is Virtual.
(c)
The magnification of the final image.
Answer to Problem 76PQ
The total magnification of the final image is
Explanation of Solution
Write the expression for the total magnification of the final image.
Here,
Write the expression for the magnification of the first lens.
Here,
Write the expression for the magnification of the second lens.
Here,
Conclusion:
Substitute the equations (V) and equation (VI) in the equation (IV) to find
Substitute
Thus, the total magnification of the final image is
(d)
Whether the final image is upright or inverted.
Answer to Problem 76PQ
The final image produced will be inverted.
Explanation of Solution
Since, the total magnification has a negative sign, the image formed will be inverted.
Want to see more full solutions like this?
Chapter 38 Solutions
Physics for Scientists and Engineers: Foundations and Connections
- A doctor examines a mole with a 15.0-cm focal length magnifying glass held 13.5 cm from the mole. (a) Where is the image? (b) What is its magnification? (c) How big is the image of a 5.00 mm diameter mole?arrow_forwardTwo converging lenses having focal lengths of f1 = 10.0 cm and f2 = 20.0 cm are placed a distance d = 50.0 cm apart as shown in Figure P35.48. The image due to light passing through both lenses is to be located between the lenses at the position x = 31.0 cm indicated. (a) At what value of p should the object be positioned to the left of the first lens? (b) What is the magnification of the final image? (c) Is the final image upright or inverted? (d) Is the final image real or virtual?arrow_forwardIn Figure P35.30, a thin converging lens of focal length 14.0 cm forms an image of the square abed, which is he = hb = 10.0 cm high and lies between distances of pd = 20.0 cm and pa = 30.0 cm from the lens. Let a, b, c. and d represent the respective corners of the image. Let qa represent the image distance for points a and b, qd represent the image distance for points c and d, hb, represent the distance from point b to the axis, and hc represent the height of c. (a) Find qa, qd, hb, and hc. (b) Make a sketch of the image. (c) The area of the object is 100 cm2. By carrying out the following steps, you will evaluate the area of the image. Let q represent the image distance of any point between a and d, for which the object distance is p. Let h represent the distance from the axis to the point at the edge of the image between b and c at image distance q. Demonstrate that h=10.0q(114.01q) where h and q are in centimeters. (d) Explain why the geometric area of the image is given by qaqdhdq (e) Carry out the integration to find the area of the image. Figure P35.30arrow_forward
- People who do very detailed work close up, such as jewelers, often can see objects clearly at much closer distance than the normal 25 cm. (a) What is the power of the eyes of a woman who can see an object clearly at a distance of only 8.00 cm? (b) What is the image size of a 1.00-mm object, such as lettering inside a ring, held at this distance? (c) What would the size of the image be if the object were held at the normal 25.0 cm distance?arrow_forwardA camera with a 50.0-mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens must the film be? (b) If the film is 36.0 mm high, what fraction of a 1.75-m-tall person will fit on it? (c) Discuss how reasonable this seems, based on your experience in taking or posing for photographs.arrow_forwardFigure P38.43 shows a concave meniscus lens. If |r1| = 8.50 cm and |r2| = 6.50 cm, find the focal length and determine whether the lens is converging or diverging. The lens is made of glass with index of refraction n = 1.55. CHECK and THINK: How do your answers change if the object is placed on the right side of the lens? FIGURE P38.43arrow_forward
- A converging lens made of crown glass has a focal length of 15.0 cm when used in air. If the lens is immersed in water, what is its focal length? (a) negative (b) less than 15.0 cm (c) equal to 15.0 cm (d) greater than 15.0 cm (e) none of those answersarrow_forward(a) What magnification is produced by a 0.150 cm-focal length microscope objective that is 0.155 cm from the object being viewed? (b) What is the overall magnification if an 8 x eyepiece (one that produces an angular magnification of 8.00) is used?arrow_forwardAn object is placed a distance of 10.0 cm to the left of a thin converging lens of focal length f = 8.00 cm, and a concave spherical mirror with radius of curvature +18.0 cm is placed a distance of 45.0 cm to the right of the lens (Fig. P38.129). a. What is the location of the final image formed by the lensmirror combination as seen by an observer positioned to the left of the object? b. What is the magnification of the final image as seen by an observer positioned to the left of the object? c. Is the final image formed by the lensmirror combination upright or inverted? FIGURE P38.129arrow_forward
- Why is the following situation impossible? Consider the lensmirror combination shown in Figure P35.55. The lens has a focal length of fL = 0.200 m, and the mirror has a focal length of fM = 0.500 m. The lens and mirror are placed a distance d = 1.30 m apart, and an object is placed at p = 0.300 m from the lens. By moving a screen to various positions to the left of the lens, a student finds two different positions of the screen that produce a sharp image of the object. One of these positions corresponds to light leaving the object and traveling to the left through the lens. The other position corresponds to light traveling to the right from the object, reflecting from the mirror and then passing through the lens. Figure P35.55 Problem 55 and 57.arrow_forwardThe left face of a biconvex lens has a radius of curvature of magnitude 12.0 cm, and the right face has a radius of curvature of magnitude 18.0 cm. The index of refraction of the glass is 1.44. (a) Calculate the focal length of the lens for light incident from the left. (b) What If? After the lens is turned around to interchange the radii of curvature of the two faces, calculate the focal length of the lens for light incident from the left.arrow_forwardAn object 1.50 cm high is held 3.00 cm from a person’s cornea, and its reflected image is measured to be 0.167 cm high. (a) What is the magnification? (b) Where is the image? (c) Find the radius of curvature of the convex mirror formed by the cornea. (Note that this technique is used by optometrists to measure the curvature of the cornea for contact lens fitting. The instrument used is called a keratometer, or curve measurer.)arrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
- Physics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning