{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MIT2_71S09_gsol1

# MIT2_71S09_gsol1 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MASSACHUSETTS INSTITUTE OF TECHNOLOGY 2.710 Optics Spring ’09 Solutions to Problem Set #1 Posted Wednesday, Feb. 18, 2009 Problem 1: Spherical waves and energy conservation In class we mentioned that the radiation from a point source can be characterized by a spherical wave in which the phase of the electric field is constant along a spherical surface (wavefront). As time evolves, the wavefronts expand outwards and thus the amplitude of the electric field at a given point changes. However, energy conservation requires the power to be constant at any radial distance from the point source after integrating over the spherical surface. The intensity of the field is defined as the power P over the area A (in Watts/m 2 ), I = P A , where for a spherical surface A = 4 πr 2 . In addition, the intensity is proportional to the square of the electric field amplitude, I = | E | 2 . Taking the square root on both sides of the previous equation, we arrive to the con- clusion that the amplitude of the electric field for a spherical wave decays like 1/r and the intensity of the wave follows an “inverse-square law”. Problem 2: Order of magnitude calculations a) From the given website, we can see that the total solar irradiance falling outside the atmosphere of the earth is approximately 1370W/m 2 . Consider a rectangular patch of 1m 2 surface area, shown in cross–section in the schematic below, illuminated by a light bulb placed 1m away from the patch’s centre. From the geometry, we can see that the light bulb, approximated as a point radiator, subtends with respect to the patch a half–angle θ = arctan . 5 1 = 0 . 4636 rad . Let’s assume that the light bulb is an isotropic radiator, and emits power P over the 4 π solid angle. Then the power reaching the 1m 2 patch is P (2 θ ) 2 / 4 π . To match the irradiance received from the sun, we require P (2 θ ) 2 4 π = 1370 ⇒ P ≈ 2 × 10 4 W = 20 kW . A commercial light bulb has a power of 100W so it requires the power of approximately 200 light bulbs at a distance as close as 1m to match the solar irradiance....
View Full Document

{[ snackBarMessage ]}

### Page1 / 9

MIT2_71S09_gsol1 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online