Intensity of Interference Pattern

[Wombats]

The Attempt at a Solution

The solution I keep coming up with is (G) , but the solution sheet says it is (C). I assumed it was (G) because since the slit width is halved the central maximum is bigger so I assumed that X would now be part of the central maximum. Therefore the intensity would be I_0.

 

[Charles Link]

I am having trouble following this question. Where in the problem does it say anything about the slit width being reduced to one half? The question seems very incomplete in their explanation for what they mean by the diffraction pattern changing. If you block or cover the slit, you get zero. I don't understand what they are trying to demonstrate. $\\$ Meanwhile though, the intensity os not constant across the central maximum. The intensity for slit width $b$ at angle $\theta$ is $I(\theta)=I_o \frac{\sin^2(\pi b \sin(\theta)/\lambda)}{(\pi b \sin(\theta)/\lambda)^2}$.

 

[Wombats]

I am having trouble following this question. Where in the problem does it say anything about the slit width being reduced to one half? The question seems very incomplete in their explanation for what they mean by the diffraction pattern changing. If you block or cover the slit, you get zero. I don't understand what they are trying to demonstrate.

Oops I guess I cropped it out of the image. Yeah it just tells you that the right half of the slit is covered. There were no numbers given so I just don't understand how to go about figuring it out. I think they are trying to demonstrate what changes in the pattern when you decrease the slit width such as the central maximum would increase, but I am not sure how to solve the problem concerning the coefficient of I_0 after the change.

 

[Charles Link]

See the addition to my post. $I_o$ will be proportional the second power of the slit width, so you need to take that into account in your calculations as well: i.e. the $I_o$ for the half-slit case will be reduced by a factor of 4.

 

[Charles Link]

And yes, I get the right answer. I can give you a couple hints: If you look at the formula I gave you, what is $\frac{b \sin(\theta)}{\lambda}$ when you are at the first zero of intensity, i.e. when $m=1$ in the formula $m \lambda=b \sin(\theta)$ ? $\\$ If you let the slit width be $b'=\frac{b}{2}$, what is the value of $\frac{b' \sin(\theta)}{\lambda}$, knowing what $\frac{b \sin(\theta)}{\lambda}$ was at that same location? $\\$ To get the complete intensity, you also need to know that the new $I_o$ that you write for the halved slit, call it $I_o',$ will have $I_o'=\frac{1}{4} I_o$. The reason for that is you have 1/2 as many Huygens sources all constructively interfering to produce the center of the central maximum. This will mean the $E$ that you compute, (a very simple completely qualitative computation), at the center of the pattern is 1/2 of the $E$ with the full slit, so the $I_o$ which is proportional to $E^2$ will be 1/4 of the $I_o$ of the full slit. $\\$ Also note, at the center of the pattern, the denominator of the intensity formula is zero, but for that case you simply take the limit as $\theta \rightarrow 0$, and the result is $I_o$.

 

[Wombats]

And yes, I get the right answer. I can give you a couple hints: If you look at the formula I gave you, what is $\frac{b \sin(\theta)}{\lambda}$ when you are at the first zero of intensity, i.e. when $m=1$ in the formula $m \lambda=b \sin(\theta)$ ? $\\$ If you let the slit width be $b'=\frac{b}{2}$, what is the value of $\frac{b' \sin(\theta)}{\lambda}$, knowing what $\frac{b \sin(\theta)}{\lambda}$ was at that same location? $\\$ To get the complete intensity, you also need to know that the new $I_o$ that you write for the halved slit, call it $I_o',$ will have $I_o'=\frac{1}{4} I_o$. The reason for that is you have 1/2 as many Huygens sources all constructively interfering to produce the center of the central maximum. This will mean the $E$ that you compute, (a very simple completely qualitative computation), at the center of the pattern is 1/2 of the $E$ with the full slit, so the $I_o$ which is proportional to $E^2$ will be 1/4 of the $I_o$ of the full slit. $\\$ Also note, at the center of the pattern, the denominator of the intensity formula is zero, but for that case you simply take the limit as $\theta \rightarrow 0$, and the result is $I_o$.

Thank you that makes sense now.

 

[Charles Link]

Thank you that makes sense now.

So what did your $I(\theta)$ look like at point $X$ for the halved-slit? As a homework helper, I'm not supposed to give the answer, but for the sake of completeness, please show the result of $I(\theta)=I_o' \,\frac{\sin^2( \pi b' \sin(\theta)/\lambda)}{(\pi b' \sin(\theta)/\lambda)^2}$ for the halved-slit at point $X$. The result is quite simple, but please show us the result you got.

 

[Wombats]

So what did your $I(\theta)$ look like at point $X$ for the halved-slit? As a homework helper, I'm not supposed to give the answer, but for the sake of completeness, please show the result of $I(\theta)=I_o' \,\frac{\sin^2( \pi b' \sin(\theta)/\lambda)}{(\pi b' \sin(\theta)/\lambda)^2}$ for the halved-slit at point $X$. The result is quite simple, but please show us the result you got.

The result I got for the coefficient was $\frac{\sin^2(\frac{ \pi}{2})}{(\pi^2)}$ which is equal to 0.1013 which was the correct answer on the solution sheet.