You have the equation -- just plug in the numbers.
1/p + 1/7 = 1/5
1/p = 1/5 - 1/7
1/p = 2/35
p = 35/2
1/p + 1/q = 1/5
1/q = 1/5 - 1/p
1/q = (p-5)/5p
q = 5p/(p-5)
The above is just algebra I, you know.
any online graphing site can do this. Try wolframalpha.com
limit as p->infinity = 5
limit as p->5 = infinity
parallel rays converge to the focal length (their origin is at infinity)
If you could give an explenation with the answers, that'd be wonderful so I actually know how to solve similar problems in the future. :) Thank you so much!
1. A convex lens with focal length f centimeters will project the image of an object on a point behind the lens. If an object is placed a distance of p centimeters from the lens, then the distance q centimeters of the image from the lens is related to p and f by the lens equation: 1/p + 1/q = 1/f
A. If the focal length of the convex lens is supposed to be 5 cm, and if the image is formed 7 cm from the lens, find the distance, p, from the lens to the object. (It’s not necessary to simplify your answer.)
B. Find an expression that gives q as a function of p, assuming that the focal length is a constant of 5 centimeters.
C. Sketch a graph of q as a function of p (i.e., q(p)), assuming that the focal length is a constant of
5 centimeters. Show any important features of the graph.
D. Find lim p->infinity q(p) and lim p->5+ q(p). What do these limits represent physically? What must happen to the distance of the image and the object?
1 answer
1 answer