Concept explainers
(a)
The position of the final image.
(a)
Explanation of Solution
Given:
The focal length of a converging lens is
The separation between the two lenses is
The distance of the object from the converging lens is
Formula used:
Draw a ray diagram to show the final image position and size.
Write the expression for thin lens equation for first image.
Here,
Write the expression for thin lens equation for second image.
Here,
Write the expression for the lateral magnification of image due to converging lens.
Here,
Write the expression for the lateral magnification due to the diverging lens.
Here,
Calculation:
Rewrite equation (1) to calculate the first image distance.
Substitute
Substitute
The distance of the object from diverging lens is calculated below.
Here,
Substitute
Rewrite equation (2) to calculate the second image distance.
Substitute
Substitute
The final distance of the image is calculated below.
Here,
Substitute
Conclusion:
Thus, the distance of the final image is
(b)
Whether the final image is real or virtual and upright or inverted.
(b)
Explanation of Solution
Given:
The distance of the second image is
The lateral magnification of the image due to the converging lens is
The lateral magnification of the image due to the diverging lens is
Formula used:
Write the expression for the overall lateral magnification for the system of two lenses.
Here, is the overall lateral magnification for the system of two lenses.
Calculation:
Substitute
Conclusion:
Thus, the image is virtual and inverted. Because the image distance,
(c)
The overall magnification of the system.
(c)
Explanation of Solution
Given:
The lateral magnification of the image due to the converging lens is
The lateral magnification of the image due to the diverging lens is
Formula used:
Write the expression for the overall lateral magnification for the system of two lenses.
Calculation:
Substitute
Conclusion:
Thus, the overall magnification for the system of two lenses is
Want to see more full solutions like this?
Chapter 32 Solutions
SAPLING PHYS SCIEN&ENG W/MULTITERM ACCE
- A 7.5x binocular produces an angular magnification of 7.50, acting like a telescope. (Mirrors are used to make the image upright.) If the binoculars have objective lenses with a 75.0 cm focal length, what is the focal length of the eyepiece lenses?arrow_forwardIf the lens of a person’s eye is removed because of cataracts (as has been done since ancient times), why would you expect an eyeglass lens of about 16 D to be prescribed?arrow_forwardA 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 thin plastic lens with index of refraction n = 1.67 has radii of curvature given by R1 = 12 0 cm and R2 = 40.0 cm. Determine (a) the focal length of the lens, (b) whether the lens Ls converging or diverging and the image distances for object distances of (c) infinity, (d) 8,00 cm, and (e) 50.0 cm.arrow_forwardA thin plastic lens with index of refraction n = 1.67 has radii of curvature given by R1 = 12 0 cm and R2 = 40.0 cm. Determine (a) the focal length of the lens, (b) whether the lens Ls converging or diverging and the image distances for object distances of (c) infinity, (d) 8,00 cm, and (e) 50.0 cm.arrow_forwardWhat is the focal length of a magnifying glass that produces a magnification of 3.00 when held 5.00 cm from an object, such as a rare coin?arrow_forward
- A lamp of height S cm is placed 40 cm in front of a converging lens of focal length 20 cm. There is a plane mirror 15 cm behind the lens. Where would you find the image when you look in the mirror?arrow_forwardThe contact-lens prescription for a nearsighted person is —4.00 D and the person has a far point of 22.5 cm. What is the power of the tear layer between the cornea and the lens if the correction is ideal, taking the tear layer into account?arrow_forwardTwo thin lenses of focal lengths f1 = 15.0 and f2 = 10.0 cm, respectively, are separated by 35.0 cm along a common axis. The f1 lens is located to the left of the f2 lens. An object is now placed 50.0 cm to the left of the f1 lens, and a final image due to light passing though both lenses forms. By what factor is the final image different in size from the object? (a) 0.600 (b) 1.20 (c) 2.40 (d) 3.60 (e) none of those answersarrow_forward
- A converging lens has a focal length of 10.0 cm. Locate the object if a real image is located at a distance from the lens of (a) 20.0 cm and (b) 50.0 cm. What If? Redo the calculations if the images are virtual and located at a distance from the lens of (c) 20.0 cm and (d) 50.0 cm.arrow_forwardAu object of height 3.0 cm is placed at 25 cm in front of a diverging lens of focal length 20 cm. Behind the diverging lens, there is a converging lens of focal length 20 cm. The distance between the lenses is 5.0 cm. Fluid the location and size of the final image.arrow_forwardFor normal distant vision, the eye has a power of 50.0 D. What was the previous far point of a patient who had laser vision correction that reduced the power of her eye by 7.00 D, producing normal distant vision?arrow_forward
- University Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStaxPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax College
- Physics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics 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 Learning