- #1

- 407

- 0

## 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 :)

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 :)