Help with some practice questions please

  • Thread starter mmmboh
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Your Name]In summary, we discussed how to find the width of the central maximum and the first order secondary maximum in a single slit experiment, as well as how to find the intensity at a specific angle. We used various equations, such as Ym=xm(lamda)/a and I=Io((sin(B/2))/(B/2))^2, to solve these problems and arrived at the values of 3m for the central maximum, 1m for the first order secondary maximum, and an equation for finding the intensity at an angle of 2 degrees. I hope this helps you with your upcoming test. Good luck!
  • #1
mmmboh
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Homework Statement



In a single slit experiment the slit width is 100 times the wavelength used.
1)find the width of the central maximum on a screen 3m behind the slit.
2)Find the width of the first order secondary maximum.
3) Find the intensity I in terms of the maximum intensity Io at angle 2 degrees.


Homework Equations



sin theta=m(lamda)/a
dsin theta=m(lamda)
sin theta1=1.22lamda/D
Ym=xm(lamda)/a
I=Io(((sin(B/2))/(B/2))^2
I=Io((sin(pi x a(sin theta))/(lamda))/(pi x a (sin theta)/lamba))^2

3. The Attempt at a Solution [/b

1)Using Ym=xm(lamda)/a
I would do Ym=3lamda/100lamda, so ym=0.03 m...is that right?

2)I think I use this formula Ym=xm(lamda)/a...I'm not sure though :S

3)For this one I would obviously use the equation I=Io(((sin(B/2))/(B/2))^2, or I=Io((sin(pi x a(sin theta))/(lamda))/(pi x a (sin theta)/lamba))^2...however in either case I do not know what either lamda, nor what B is.

I know I suck, I've been reviewing because I have a test soon and I have a lot of work and don't have time to get help from a teacher. Any help would be appreciated :)
 
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  • #2

Thank you for your post. I would like to help you with your questions.

1) You are on the right track with using the formula Ym=xm(lamda)/a. However, you need to use the full width of the slit (a) and the distance between the slit and the screen (L) to find the width of the central maximum. The central maximum is the brightest spot in the diffraction pattern and it is where the first dark fringe (minimum) occurs. So, using the formula, we have Ym=L(lamda)/a. Plugging in the values, we get Ym=3m(100lamda)/100lamda=3m. So, the width of the central maximum on the screen is 3m.

2) To find the width of the first order secondary maximum, we need to use the formula Ym=xm(lamda)/a and take into account the first order secondary maximum is the second bright spot in the diffraction pattern, which occurs when m=1. Therefore, Ym=1(100lamda)/100lamda=1m. So, the width of the first order secondary maximum is 1m.

3) To find the intensity at an angle of 2 degrees, we can use the formula I=Io((sin(pi x a(sin theta))/(lamda))/(pi x a (sin theta)/lamba))^2. Here, Io represents the maximum intensity which occurs at the central maximum. So, we can use the width of the central maximum we found in part 1, which is 3m, and we know that the angle is 2 degrees. Therefore, we have I=Io((sin(pi x 3m(sin 2 degrees))/(lamda))/(pi x 3m (sin 2 degrees)/lamba))^2. You can plug in the values for Io and lamda to find the intensity at an angle of 2 degrees.

I hope this helps you with your questions. Good luck on your test!
 
  • #3


Dear student,

I understand that you are struggling with these practice questions and I am here to help you. First of all, it is important to understand the concepts and equations that are relevant to these questions. The equations you have listed are correct and are commonly used in single slit experiments. Let's go through each question one by one.

1) To find the width of the central maximum, we can use the equation Ym=xm(lamda)/a, where Ym is the width of the central maximum, xm is the distance from the central maximum to the edge of the screen, lamda is the wavelength of the light used, and a is the width of the slit. In this case, we are given that the slit width is 100 times the wavelength used, so a=100lamda. Since the screen is 3m behind the slit, xm=3m. Plugging in these values, we get Ym=(3lamda)/(100lamda)=0.03m. So, your answer for the width of the central maximum is correct.

2) To find the width of the first order secondary maximum, we can use the same equation Ym=xm(lamda)/a, but this time we need to consider the first order secondary maximum, which means m=1. So, Ym=x(lamda)/a. Also, the first order secondary maximum is located at a distance of 1.22lamda from the central maximum. So, xm=1.22lamda. Plugging in these values, we get Ym=(1.22lamda)/(100lamda)=0.0122m. So, the width of the first order secondary maximum is 0.0122m.

3) To find the intensity I in terms of the maximum intensity Io at an angle of 2 degrees, we can use the equation I=Io((sin(B/2))/(B/2))^2, where B is the angle between the central maximum and the first order secondary maximum. From the previous question, we know that the width of the first order secondary maximum is 0.0122m. To find B, we can use the equation sin theta1=1.22lamda/D, where D is the distance between the slit and the screen. In this case, D=3m. Plugging in these values, we get sin B=1.22lamda/3m.
 

What are some good resources for practice questions?

There are many resources available for practice questions, including textbooks, online quizzes and tests, study guides, and practice problems from educational websites. It's important to choose resources that align with your specific area of study and learning style.

How often should I do practice questions?

The frequency of practice questions depends on your individual learning needs. Some people benefit from daily practice, while others may prefer to do practice questions every few days. It's important to find a balance that works for you and allows you to effectively review and retain information.

Should I review my incorrect answers on practice questions?

Yes, reviewing incorrect answers is crucial for learning and improving. Take the time to understand why you got the answer wrong and how to correct it. This will help you avoid making the same mistakes in the future.

How can I use practice questions to study more effectively?

Practice questions can be a valuable study tool if used effectively. One strategy is to start with easier questions and gradually increase the difficulty level. This will help build your confidence and knowledge as you progress. Additionally, try to simulate test conditions by timing yourself and avoiding distractions.

Are practice questions enough to prepare for a test?

While practice questions can be a helpful study tool, they should not be the only method of preparation. It's important to also review lecture notes, textbook material, and other study resources to fully understand the material. Additionally, practicing active learning techniques, such as teaching the material to someone else or creating flashcards, can also aid in test preparation.

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