1/p + 1/q = 1/f
M = -q/p
where p is the distance of the object from the lens, q is the distance of the image from the lens, M is the magnification of the object. If M is positive the image is virtual and inverted; if M is negative the image is real and upright (erect).
f is positive for a converging lens:
1/10 + 1/q = 1/15
Solve for q, and then calculate M, and determine the rest of the answers. Sorry, I can't draw a ray diagram for you, but here are the rules:
To draw these ray diagrams, we will have to recall the three rules of refraction for a double convex lens:
Any incident ray traveling parallel to the principal axis of a converging lens will refract through the lens and travel through the focal point on the opposite side of the lens.
Any incident ray traveling through the focal point on the way to the lens will refract through the lens and travel parallel to the principal axis.
An incident ray that passes through the center of the lens will in effect continue in the same direction that it had when it entered the lens.
A birthday candle 4 cm high is 10 cm from a converging lens whose focal length is 15 cm. Use a ray diagram on a suitable scale to find the location of the image, its height, and whether it is erect or inverted, real or virtual.
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