The moon's diameter is 3.48 ×10^6 m, and its mean distance from the earth is 3.85 ×10^8 m . The

The moon's diameter is 3.48 ×10^6 m, 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?

Key Concepts

Thin Lens Equation

The thin lens equation (1/f = 1/d? + 1/d?) is a fundamental formula in geometric optics that relates the focal length of a lens to the distances of the object and its image. It serves as the basis for determining how a lens will form an image given the position of the object and is crucial for calculating image distances in both photography and projection systems.

Magnification in Optical Systems

Magnification is defined as the ratio of the size of the image to the size of the object, and in a lens system, it can be approximated as m = d?/d?. This concept allows one to calculate how large (or small) an image will appear relative to the original object, playing a key role in designing optical setups for cameras and projectors.

Similar Triangles in Geometric Optics

The principle of similar triangles underlies the derivation of the magnification formula in optics. By considering the geometry of light rays passing through a lens, one can relate the dimensions of the object to the dimensions of the image, a method that is essential for understanding and solving imaging problems.

Unit Conversion in Optical Calculations

Consistent unit conversion is critical in optical calculations to maintain accuracy. Optical problems often involve different units such as meters and millimeters, and ensuring all quantities are in compatible units is necessary for correctly applying formulas like the thin lens equation and magnification formulas.

Transcript

00:01 So for part a, we start with 1 over do plus 1 over d .i is equal to 1 over f.

00:09 Now, the distance of the moon from the earth is 3 .85 times, so this is do, is 3 .85 times 10 to the power 8 meters.

00:21 What that means for 1 over do is that it's almost zero compared to the other distances.

00:27 So we can then say that d -i is approximately the same as f.

00:32 And f is given as 50 millimeters or 0 .0 meters, 0 .05 meters.

00:44 Now, the magnification, we want to find the distance of the image, so the height of the image.

00:51 So h -i over h -o, that is equal to the magnitude of d -i over d -o.

00:59 So we have h -i is what we're trying to find out.

01:05 H -o would be the diameter of the moon, and that is equal to 3 .48 times 10 to the power 6 meters.

01:18 And that is equal to the distance of the image, which we just figured out is 0 .05 meters, divided by the distance of the object, which is the distance between earth and the moon.

01:29 So that's 3 .85.

01:30 Times 10 to the power 8 meters.

01:34 So solving this equation for h .i gives h .i is equal to 4 .52 times 10 to the negative 4 meters.

01:43 So the diameter of the moon on the screen would be 4 .52 times 10 to the negative 4 meters.

01:50 Now for part b, the same object, the same image now becomes the object...

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