Concept explainers
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′, h′b represent the distance from point b′ to the axis, and h′c represent the height of c′. (a) Find qa, qd, h′b, and h′c. (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
where h′ and q are in centimeters. (d) Explain why the geometric area of the image is given by
(e) Carry out the integration to find the area of the image.
Figure P26.38
(a)
The values of given parameters.
Answer to Problem 38P
The values of the given parameters are
Explanation of Solution
Write the mirror equation for the side a.
Here
Rewrite (I) in terms of
Similarly, write the image distance for side d.
Write the equation for magnification for side b
Here
Rewrite (IV) in terms of
Similarly,
Conclusion:
Substitute
Substitute
Substitute
Substitute
The values of the given parameters are
(b)
Sketch of the image.
Answer to Problem 38P
The image has been drawn
Explanation of Solution
Conclusion:
The image has been drawn
(c)
Prove the given equation.
Answer to Problem 38P
The equation has been proved.
Explanation of Solution
Write the equation for magnification
Rewrite (VII) in terms of
Write the mirror equation in terms of
Substitute (IX) in (VIII)
Conclusion:
Substitute
Hence proved.
(d)
Geometric area of the image.
Answer to Problem 38P
The geometric area of the image is given by
Explanation of Solution
The integral sign suggests that the areas of the small regions are added up to get the whole area. The
Conclusion:
The geometric area of the image is given by
(e)
Area of the image.
Answer to Problem 38P
The area is
Explanation of Solution
Write the equation for area
Substitute (XI) in (XII)
Conclusion:
The area is
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Chapter 26 Solutions
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