The moon's diameter is 3.48 × 10^6m, and its mean distance...

The moon's diameter is $3.48 \times 10^{6} \mathrm{m},$ and its mean distance from the earth is $3.85 \times 10^{8} \mathrm{m} .$ The moon is being photographed by a camera whose lens has a focal length of 50.0 $\mathrm{mm}$ . (a) Find the diameter of the moon's image on the slide film. (b) When the slide is projected onto a screen that is 15.0 $\mathrm{m}$ from the lens of the projector $(f=110.0 \mathrm{mm})$ what is the diameter of the moon's image on the screen?

Transcript

00:01 Suppose we have a camera that photographs the moon and as a lens with a focal length of 50 millimeter.

00:09 And we know that the distance of the moon from the earth is 3 .85 times 10 race to 8 meters.

00:19 So we will denote this as a distance of the object from the nemes.

00:27 And the diameter of the moon is 3 .48 times 10 to 6 meters.

00:33 So we will denote a size of the object.

00:39 Now, question a, one as to determine the size of the image on the field, so the diameter of the moon image in the field.

00:49 So, we will determine this using the formula for multiplication.

00:54 So the magnification equals to the ratio of the size of the image over the size of the object, which is also equals to the negative of the ratio of the distance of the image over the distance of the object.

01:09 Then isolating the sizes of the image at the left side of the question, since this is what mean the other equation comes like this.

01:16 So in order to hold this problem, we will have to determine the distance of the image first, since the size of the object is given and the distance of the object is also given.

01:29 So we can determine this using the thin lens equation, which is this, since it relates the distance of the image, distance of the object, and the focal length, which we also have from the problem.

01:47 Then isolating the distance of the image to the left side since this is what we needed, then our equation becomes like this.

02:03 Then solving for the distance of the image, we have one over the focal length.

02:08 If converted to meters, we just have to multiply 10 raise to negative 3.

02:14 So the distance of the image is 0 .85 distance of the, this is the object i mean.

02:22 So 3 .85 times 10 raise to 8 meters.

02:26 Of the object which is the distance of the moon from earth.

02:29 Then solving we can find that the distance of the image is 0 .05 meters.

02:36 Then we can solve for the size of the image...