The sun is 150,000,000 km from earth; its diameter is 1,400,000 km . A student uses a 4.0-cm- di

The sun is 150,000,000 km from earth; its diameter is...

The sun is $150,000,000 \mathrm{km}$ from earth; its diameter is $1,400,000 \mathrm{km} .$ A student uses a $4.0-\mathrm{cm}-$ diameter lens with $f=10 \mathrm{cm}$ to cast an image of the sun on a piece of paper. a. Where should the paper be placed to get a sharp image? b. What is the diameter of the image on the paper? c. The intensity of the incoming sunlight is $1050 \mathrm{W} / \mathrm{m}^{2} .$ What is the power of the light captured by the lens? d. What is the intensity of sunlight in the projected image? Assume that all of the light captured by the lens is focused into the image.

The sun is $150,000,000 \mathrm{km}$ from earth; its diameter is $1,400,000 \mathrm{km} .$ A student uses a $4.0-\mathrm{cm}-$ diameter lens with $f=10 \mathrm{cm}$ to cast an image of the sun on a piece of paper.
a. Where should the paper be placed to get a sharp image?
b. What is the diameter of the image on the paper?
c. The intensity of the incoming sunlight is $1050 \mathrm{W} / \mathrm{m}^{2} .$ What is the power of the light captured by the lens?
d. What is the intensity of sunlight in the projected image? Assume that all of the light captured by the lens is focused into the image.

Key Concepts

Thin Lens Equation

The thin lens equation, given by 1/f = 1/do + 1/di, is a fundamental concept in geometric optics that relates the focal length of a lens (f) to the object distance (do) and the image distance (di). This equation is crucial in determining where an image will be formed, especially when dealing with distant objects where the approximation can simplify the analysis.

Image Formation and Focal Length

Image formation in lenses relies on focusing light rays that come from a distant object to a point determined by the lens’ focal length. When the object is extremely far away, like the sun, the image is formed at or near the focal plane of the lens. This understanding helps in setting up experiments where a clear image is required.

Magnification

Magnification in the context of lens optics is defined as the ratio of the image size to the object size and is related to the distances by the formula m = di/do. For objects at a great distance, the magnification is very small, resulting in a significantly reduced image size compared to the actual object size.

Aperture and Light Collection

The aperture of a lens, determined by its diameter, defines the area through which light is admitted. The amount of light collected by the lens is proportional to this area, which is essential for calculating the total power of light captured from a source with a known intensity.

Intensity and Power of Light

Intensity is defined as the power per unit area (W/m²) and plays a key role in determining the brightness of an image. When a lens concentrates light, the power captured is the product of the incoming light intensity and the lens’s area. Understanding this helps in analyzing how light energy is conserved or redistributed when focused into a smaller area.

Transcript

00:01 In this question, it is about one student using a 4 cm diameter lens with focal length 10 cm to cast an image of the sun on a piece of paper.

00:11 So the situation is kind of like, so you have a lens, converging lens because the focal length is positive, and then you have a piece of paper and then to see the sun, so sign is very far away, so you produce parallel rays, would hit the lens, then you get focus on to the paper.

00:42 Okay, so we have four parts in this question.

00:51 In part a, where should the paper be placed to get a sharp image? so in part a, we know that s is 150 million km so 1 .5 times 10 to the 11 meters our f is 10 cm which is 0 .1 meters and using the thin length equation 1 over s plus 1 over s prime equals to 1 over f as prime is equal to 0 .1 because s is very large so one over as is approximately zero so s prime will be 0 .1 meters okay so the paper the paper should be placed at ten cm from the lens okay so ten seem from the lens is the answer in part b i want to find a diameter of the image on the paper.

02:18 So to find a diameter, we can use h prime over h equals to mod m, which is an absolute value of the magnification, which is the absolute value of minus s prime over s.

02:43 So h prime is equal to minus s prime over s absolute value times h okay so each is the diameter and we know that s prime is 0 .1 meters so 0 .1 divide by 1 .5 times 10 to the 11.

03:06 H is the diameter.

03:07 It will be 1 .4 times 10 to the 9.

03:12 And calculate this you get 9 .33 times 10 to the minus 4 meters.

03:23 Okay so this is the answer for this question for part b...

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