Recent content by fubag

just curious then, the formula for constructive interference in a thin film is given by: 2nd = (m+0.5)lambda and the destructive interference in a thin film is then given by: 2nd = m(lambda) ?

ok so using that... the thinnest it can be is when m = 1, and therefore d = (lambda) / (2.66) ?

ok i actually solved it now found my mistake thanks!

just a hunch but I thought about plugging in some numbers given in the problem: 2(1.33)d = (m + 0.5)(lamda) d = ((m + 0.5)lamda)/(2.66) if we move the 0.5 down, and assign m = 3 I get choice B = (3*lambda)/(5.32) does this logic make sense? it seems to fit the equation?

ok so I find (theta_min) = (550 * 10^-9) / (.36m), using a as distance between the two points. then solve for D = (1.22(550 * 10^-9)) / (theta_min) ? This gives me .44m which makes no sense.

ok still stuck on this one... I understand I am using Rayleigh's criterion formula where (theta_min) = lamda / a . So I think I have to take a = .36m, since it is now the width of the slit or something. and using the properties that tan (a) = opposite/adjacent, i want to solve for opposite...

ok thanks a lot!

ok i solved the first part, but having difficulty with the 2nd part. When red light is directed normal to the surface of the soap film, the soap film appears black in reflected light. What is thinnest the film can be? Let lambda_0 be the wavelength of red light in air. a. {lambda_0}/{2} b...

i think the answers are b, and f could someone please verify?

ok I got that part Part 2: Assuming that the screen is sufficiently bright, at what distance can you no longer resolve two pixels on diagonally opposite corners of the screen, so that the entire screen looks like a single spot? Note that the size (0.360 meters) quoted for a monitor is the...

ok so I used theta_min = atan(281micrometers/1.3m) and solve for D? This gave me D = 3.10 mm. Is this correct?

lamda would change by a factor of decreasing 1.33 so the width would decrease therefore?

[SOLVED] Interference in Soap films Homework Statement The sketch View Figure shows a thin film of soapy water of uniform thickness t and index of refraction n = 1.33 suspended in air. Consider rays 1 and 2 emerging from the film: Ray 1 represents the reflected wave at the air-film...

[SOLVED] Resolving pixels on a computer screen Homework Statement A standard 14.16-inch (0.360-meter) computer monitor is 1024 pixels wide and 768 pixels tall. Each pixel is a square approximately 281 micrometers on each side. Up close, you can see the individual pixels, but from a distance...

[SOLVED] Single-Slit Diffraction Homework Statement You have been asked to measure the width of a slit in a piece of paper. You mount the paper 80.0 centimeters from a screen and illuminate it from behind with laser light of wavelength 633 nanometers (in air). You mark two of the intensity...