An object is placed 4.0 cm to the left of a converging...

An object is placed 4.0 cm to the left of a converging lens which has a focal length of 6.0 cm. The image is 1.0 cm tall. (a) Is the image real or virtual? (b) Use the thin lens formula to calculate the image distance, s'. Show your work. (c) Draw a principle-ray diagram for this situation. Your diagram must be to scale, and you must use three rays to locate the image. (You may need to scale down your diagram in order for it to fit on your paper.)

Transcript

00:01 Hi there, so for this problem, we are told that an object is placed 4 centimeters to the left of a converging lens.

00:07 So the object distance is given, and we are going to call that do, and that is equal to 4 centimeters.

00:16 Now, with that set, the converging lens has a focal length that is given, and that focal length is equal to 6 centimeters.

00:27 And the image is one centimeters tall.

00:34 So we are also given its height, the height of the image, one centimeter.

00:40 So for part a of this problem, we are asked about if the image is real or virtual.

00:49 Now, in this case, since the object is placed at four centimeters, which is less, than the focal length.

01:09 So because of that, because the object distance is less than the focal length, then we can conclude that the image in this case is going to be a virtual image, because that image is going to form in the same side as the object.

01:30 Now, with that said, we pass to par b of this problem, and we are asked about to use the thin lens formula to calculate the image distance s -prime.

01:44 Now, for that we use the fact that the inverse of the focal length is equal to the inverse of the object distance plus the inverse of the image distance s -prime.

01:54 So solving for the image distance s -prime, that is going to be equal to the inverse of the focal length minus the inverse of the object distance, and all of that elevated to the minus 1.

02:07 So now we simply substitute the values in here.

02:11 So we will have the focal length that is 6 centimeters minus the image distance that in this case is 4 centimeters, and all of that to the minus 1.

02:21 So using our calculator, we obtain a value of.

02:31 So that will be 1 divided by 6.

02:36 Minus 1 divided by 4 and all of that elevated to the minus 1.

02:41 So we obtain a value of minus 12 centimeters.

02:49 So that's a solution for part b of this problem.

02:59 Now, for par c, we are asked to draw a principal rate diagram for this situation.

03:06 Your diagram must be to a scaled and you must use three rates to locate a the image.

03:16 So let's start by drawing the object.

03:20 So let me just draw it in blue...