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Concept explainers
In 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′,
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 P35.30
(a)
![Check Mark](/static/check-mark.png)
The position of the given constants.
Answer to Problem 30P
The values of
Explanation of Solution
Given info: The focal length of the lens is
Formula to calculate the image of any object in a thin lens is,
Here,
Substitute
Here,
Substitute
Substitute
Here,
Substitute,
Formula to calculate the height of the corresponding image point
Here,
Substitute
For calculating the image height corresponding to
Substitute
Conclusion:
Therefore, the values of
(b)
![Check Mark](/static/check-mark.png)
To draw: The sketch of the ray diagram.
Answer to Problem 30P
The sketch of the ray diagram is,
Explanation of Solution
The image of the square
Figure (1)
(c)
![Check Mark](/static/check-mark.png)
To show: The relation,
Answer to Problem 30P
The relation between height of the image and the image distance is
Explanation of Solution
the focal length of the given lens is
Here,
From the lens,
Here,
Substitute
Substitute
Conclusion:
Therefore the relation between height of the image and the image distance is
(d)
![Check Mark](/static/check-mark.png)
To write: The explanation that the geometric area of image is
Answer to Problem 30P
The geometric area of a image is explained by the integral
Explanation of Solution
Given info: The geometric area of a image is,
Here,
From equation (10) the integral sums up the small areas of region covered by the image itself. The height of the small regions is
Therefore area of that small region is
Therefore the integration from
Conclusion:
Therefore, the geometric area of the image is given by integral
(e)
![Check Mark](/static/check-mark.png)
The geometric area of the image.
Answer to Problem 30P
The geometric area is
Explanation of Solution
Given info: The geometric area of the image is given by the integral,
From equation (9) substitute
Integrate the above equation with respect to
Conclusion:
Therefore, the geometric area of the image is
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Chapter 35 Solutions
Physics for Scientists and Engineers with Modern Physics
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