Note that when we are solving these problems, we will use Do instead of Do to represent object distance, and Di instead of Di to represent image distance. It is hoped that you will continue to consider the o and i as subscripts and use them as such in your own work. We will also use the exponent -1 instead of writing out the inverse fraction in many cases. This is to save key strokes!
Problem 1:
A convex lens with a focal length of 15 cm is placed 25 cm from a lighted object. At what distance will an image be formed? Will it be real?
Givensf = 15 cm
Do = 25 cmUnknown
Di = ?Equation
1/f = 1/Do + 1/DiSolving
1/Di = 1/f - 1/Do = 1/15 cm - 1/25 cm = 0.0267 cm-1Di = (0.0267 cm-1)-1 = 37.5 cm
Because the object is further from the lens than the focal point, the image will be real. Also notice that the image distance is a positive one. This means that it is on the other side of the lens and is therefore a real image. Finally, the image distance is larger than the object distance. Therefore the image is larger than the object. (Remember that magnification is simply the ratio of image distance to object distance, Hi/Ho.)
Problem 2:
An object is placed 25 cm from a convex lens which has a focal length of 10 cm. At what distance will the image be formed? Will it be real? Will it be enlarged or reduced in size?
Givensf = 10 cm
Do = 25 cmUnknown
Di = ?Equation
1/f = 1/Do + 1/DiSolving
1/Di = 1/f - 1/Do = 1/10 cm - 1/25 cm = 0.06 cm-1Di = (0.06 cm-1)-1 = 16.67 cm
Because the object is further from the lens than the focal point, the image will be real. Also notice that the image distance is a positive one. This means that it is on the other side of the lens and is therefore a real image. Finally, the image distance is less than the object distance. Therefore the image is smaller than the object. (Remember that magnification is simply the ratio of image distance to object distance, Hi/Ho.)
Problem 3:
Where should one place a lighted object so that the final image is 4 meters away from a lens having a focal length of 20 cm? Will the final image be enlarged or reduced?
Givensf = 20 cm
Di = 400 cmUnknown
Do = ?Equation
1/f = 1/Do + 1/DiSolving
1/Do = 1/f - 1/Di = 1/20 cm - 1/400 cm = 0.0475 cm-1Di = (0.0475 cm-1)-1 = 21.05 cm
Because the image is further from the lens than the focal point, the image will be real. The image distance is much greater than the object distance. Therefore the image is enlarged. (Remember that magnification is simply the ratio of image distance to object distance, Hi/Ho.) This is similar to having a slide projector. In fact, where wouldyou place the slide you need to project? Yes, just a little ways from the focal point of the lens.
Problem 4:
We have an object that's very far away (100 meters) compared to the 15-cm focal length of our convex lens. Where will the image be formed? What are its characteristics?
Givensf = 15 cm
Do = 104 cmUnknown
Di = ?Equation
1/f = 1/Do + 1/DiSolving
1/Di = 1/f - 1/Do = 1/15 cm - 1/104 cm = 0.06657 cm-1Di = (0.066657 cm-1)-1 = 15.02 cm
Note that the image distance is very close to being the same as the focal length. Objects that are very far away from convex lenses will form images that are at or very close to the focal point. This is helpful so that we can design cameras that clearly focus objects at a wide range of distances, as long as they are relatively far from the lens.
Because the object is further from the lens than the focal point, the image will be real. The image distance is a positive one which means that it is on the other side of the lens and is therefore a real image. Finally, the image distance is much less than the object distance. Therefore the image is smuch maller than the object. (Remember that magnification is simply the ratio of image distance to object distance, Hi/Ho.)
Another way of thinking about this situation is to consider that an object very far away presents the lens with light rays that are almost parallel to one another. (The angle they make when striking the lens is very small.) Therefore the place they would be brought together must be close to the focal point.
Problem 5:
We have a convex lens with a focal length of 15 cm. We wish to place an object so that the final image will be 1/3 the size of the object. Where should we place that object?
Givensf = 15 cm
Hi = 1/3 HoUnknown
Do = ?Equations
Hi/Ho = Di/Do1/f = 1/Do + 1/Di
Solving
Hi / Ho = (1/3 Ho) / Ho = 1/3 = Di / DoTherefore Di = 1/3 Do
1/f = 1/Do + 1/Di = 1 / Do + 1 / (1/3 Do) = 1/Do + 3/Do = 4/Do
4/Do = 1/f ..... Do = 4 f = 4 x 15 cm = 60 cm
Proof: 1/Di = 1/f - 1/Do = 1/15 cm - 1/60 cm = 0.05 cm-1
Di = (0.05 cm-1)-1 = 20 cmHi/Ho = Di/Do = 20 cm / 60 cm = 1/3 QED
Problem 6:
We place an object 20 cm from a convex lens with a focal length of 25 cm. Where is the image formed?
Givensf = 25 cm
Do = 20 cmUnknown
Di = ?Equation
1/f = 1/Do + 1/DiSolving
1/Di = 1/f - 1/Do = 1/25 cm - 1/20 cm = -0.01 cm-1Di = (-0.01 cm-1)-1 = -100 cm
We did get an answer. We worked carefully, but the answer turned out negative. At this point we may not have a way to think about negative answers, but with a non-positive answer we conclude that the image is likely not real. To learn more about what this negative answer means, go to the section on VIRTUAL IMAGES.
We hope this set of problems is useful in helping you with your problem-solving using convex lenses. Remember to use the GUESS formula in all of your work. Include units in all of your work, too, and be sure to check your final answers.
GUESS FORMULA:
G = Givens - write them all down with symbols, numbers and unitsU = Unknown - write down the unknown with a question mark
E = Equation - write down the equation or equations that apply in the problem
solve the equation(s) for the desired unknownS = Substitute - substitute the given information into the solved equation
S = Solve - solve numerically, paying attention to units
identify the final answer by circling, boxing, or underlining it clearly
check the final answer to be sure it makes sense