Question

A single vertical slit of width 24000 nm is located 2.10 m from a screen. The intensity pattern below is observed. The distances on the screen can be measured from the scale. Each tick mark is equal to 1 cm. Intensity is displayed along the y-axis. What is the wavelength of the light?

My attempt at the solution:

1cm = 0.01m

From the intensity pattern diagram, it appears that the first m of diffraction = 1 occurs at the fourth tick mark (which I interpret as 0.04m from the tick mark at the central max).

tan(theta) = y/L
= 0.04m/2.10m
= 0.019047619
= 0.0190
theta turns out to be 1.091216225deg which is small, so I used the "small angle approximation" in which tan(theta) = sin(theta) = 0.0190 approximately.

asin(theta) = (m)(lambda)
(24000*10^-9)(0.0190) = (1)(lambda)
lambda = 4.57*10^-7 m

That's the value I got for lambda but apparently it's incorrect. Could someone please help me figure out what I did wrong?

Answer 1

are you certain the distance from the center to the first minium is .04m?

Answer 2

[Replace each space with a dot(.)] Link to image related to this question-> lh3 googleusercontent com/TjcBG60hMmLnA7rwTp0hIb4SzI2gqZtebe8g3cjQ1rf8xd0HXulzCNZTrTrVzw0zPyNC=s147

Answer 3

Oh wait, that might not have been posted properly. Just put a dot after "Ih3" and "googleusercontent". I needed to show you the image because I feel that is the issue, that I may not have correctly interpreted the distance of the first m of diffraction.

Answer 4

Nevermind, apparently that distance value was 0.043 which I was not certain before because it was based on more of an estimation rather than a definite value. So all good now!