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?
are you certain the distance from the center to the first minium is .04m?
[Replace each space with a dot(.)] Link to image related to this question-> lh3 googleusercontent com/TjcBG60hMmLnA7rwTp0hIb4SzI2gqZtebe8g3cjQ1rf8xd0HXulzCNZTrTrVzw0zPyNC=s147
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.
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!
Well, it seems like you're having a "wavelength of light" bulb moment! Let's shed some comedic light on this situation.
From what I understand of your calculations, you used the formula asin(theta) = m(lambda), where m represents the order of the diffraction pattern. However, in this case, it seems like you assumed m = 1, which might be causing your discrepancy.
But, don't worry! We can enlighten your approach a bit. Instead of assuming m = 1, let's first figure out what m actually represents. Looking at the intensity pattern, can you spot any other peaks or troughs apart from the first one?
Remember, when you have a single vertical slit, you'll usually observe multiple bright spots, which means there are multiple maxima in the pattern. So, to calculate the correct wavelength, we need to consider the number of peaks or troughs you observed on the screen.
Once you identify the correct value for m, you can plug it into the formula and get a more accurate wavelength.
Keep up the bright work, and don't worry, we'll get to the bottom of this conundrum together!
To find the wavelength of the light, you used the equation asin(theta) = m(lambda), where asin(theta) is the angle of diffraction, m is the order of the maximum, and lambda is the wavelength of the light.
You correctly determined the value of theta as 1.091216225 degrees using the small angle approximation. However, there is a small mistake in your calculation.
You correctly calculated the value of tan(theta) as 0.0190, but instead of using asin(theta), you should use the inverse tangent function to find the value of theta.
theta = arctan(tan(theta)) = arctan(0.0190) = 1.091213759 degrees (approximately).
Now, to find the wavelength lambda, you will use the correct value of theta:
lambda = (w * sin(theta)) / m
where w is the slit width and m is the order of the maximum.
Plugging in the values:
lambda = (24000 * 10^-9 m) * sin(1.091213759 degrees) / 1
Calculating this expression, you would get the wavelength lambda as approximately 4.04 * 10^-7 m (or 404 nm).
So, the correct value for the wavelength of the light is approximately 404 nm.