Answer:
A. Using the lens equation, 1/p + 1/q = 1/f, and substituting f = 5 cm and q = 7 cm, we can solve for p:
1/p + 1/7 = 1/5
Multiplying both sides by 35p, we get:
35 + 5p = 7p
Simplifying and rearranging, we get:
2p = 35
Therefore, the distance from the lens to the object, p, is:
p = 35/2 cm
B. Solving the lens equation, 1/p + 1/q = 1/f, for q, we get:
1/q = 1/f - 1/p
Substituting f = 5 cm, we get:
1/q = 1/5 - 1/p
Multiplying both sides by 5qp, we get:
5p = qp - 5q
Simplifying and rearranging, we get:
q = 5p / (p - 5)
Therefore, the expression that gives q as a function of p is:
q = 5p / (p - 5)
C. Here is a sketch of the graph of q(p):
The graph is a hyperbola with vertical asymptote at p = 5 and horizontal asymptote at q = 5. The image distance q is positive for object distances p greater than 5, which corresponds to a real image. The image distance q is negative for object distances p less than 5, which corresponds to a virtual image.
D. Taking the limit of q as p approaches infinity, we get:
lim q(p) = 5
This represents the horizontal asymptote of the graph. As the object distance becomes very large, the image distance approaches the focal length of the lens, which is 5 cm.
Taking the limit of q as p approaches 5 from the positive side, we get:
lim q(p) = -infinity
This represents the vertical asymptote of the graph. As the object distance approaches the focal length of the lens, the image distance becomes infinitely large, indicating that the lens is no longer able to form a real image.
In order for the lens to form a real image, the object distance p must be greater than the focal length f. When the object distance is less than the focal length, the lens forms a virtual image.