Concept explainers
(a)
The focal length of eyepiece.
(a)
Explanation of Solution
Given:
Magnifying power of microscope is
The angular magnification of eyepiece is
The distance of objective lens from eyepiece is
Formula used:
Write expression for angular magnification of eyepiece.
Here,
Calculation:
Substitute
Conclusion:
Thus, the focal length of eyepiece is
(b)
The location object so that it is in focus for normal eye.
(b)
Explanation of Solution
Given:
Magnifying power of microscope is
The angular magnification of eyepiece is
The distance of objective lens from eyepiece is
Formula used:
Write expression for angular magnification of eyepiece.
Here,
Write expression for image distance.
Write expression for magnifying power of microscope.
Rearrange above expression for
Substitute
Substitute
Rearrange above expression for
Calculation:
Substitute
Substitute
Conclusion:
Thus, the object is
(c)
The focal length of objective lens.
(c)
Explanation of Solution
Given:
Magnifying power of microscope is
The angular magnification of eyepiece is
The distance of objective lens from eyepiece is
Formula used:
Write expression for angular magnification of eyepiece.
Here,
Write expression for image distance.
Write expression for magnifying power of microscope.
Rearrange above expression for
Substitute
Substitute
Rearrange above expression for
Write expression for lens equation for objective lens.
Calculation:
Substitute
Substitute
Substitute
Conclusion:
Thus, the focal length of objective lens is
Want to see more full solutions like this?
Chapter 32 Solutions
Physics For Scientists And Engineers Student Solutions Manual, Vol. 1
- Two 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_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_forwardTwo converging lenses having focal length of f1 = 10.0 cm and f2 = 20.0 cm are placed d = 50.0 cm apart, as shown in Figure P23.44. The final image is to be located between the lenses, at the position x = 31.0 cm indicated. (a) How far to the left of the first lens should the object be positioned? (b) What is the overall magnification of the system? (c) Is the final image uptight or inserted? Figure P23.44arrow_forward
- In Figure P26.38, a thin converging lens of focal length 14.0 cm forms an image of the square abcd, which is hc = 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 P26.38arrow_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_forwardShow that the magnification of a thin lens is given by M = di/do (Eq. 38.6). Hint: Follow the derivation of the lens makers equation (page 1233) and start with a thick lens.arrow_forward
- How far should you hold a 2.1 cm-focal length magnifying glass from an object to obtain a magnification of 10 x ? Assume you place your eye 5.0 cm from the magnifying glass.arrow_forwardIn Figures CQ36.11a and CQ36.11b, which glasses correct nearsightedness and which correct farsightedness?arrow_forwardA 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_forward
- College PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax CollegePrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningUniversity Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStax
- Physics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningPhysics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning