Physics 3: Images, Interference and Difraction

[Kiwithepike]

So three problems I am stuck on.
1.) You have a converging lens with n=1.5,for a symmetric lens so the two lens have same magnitude, what should the radius of curvature be so the focal length is 10cm. I know 1/f = (n-1)[1/r1-1/r2] so that is 1/0.10m =(1.5-1)[1/r1-1/r2] where I am lost is in the r1 and r2. would r2 be -r1 so we get 2r?

2)A 3cm tall object is located 10cm in front of a converging lense with a focal length of 15cm. What is the magnification of the imaged formed? |m|= h'/h and m =-i/p h is 3cm and f=15cm and p=10cm which gives a i =-30cm so would the image m be 3cm? or am i missing something?

3.) The minimum width of a slit for single diffraction to produce an interference pattern is? no min, lamda, lambda/2, or 2 lamda. since its sin=lambda/a where a is the slit width as long as a>0 there's no limit?

 

[Simon Bridge]

1. both sides have the same radius - so there are only two possibilities.
Try both and see which makes sense.

2. check by sketching the ray diagram

3. this one will depend on your course.
strictly - there is no maximum, you are right. You also get diffraction for the limit of zero width.
diffraction can occur around an edge - so the "slit width" there is infinite.
however - your course may be talking about a specific kind of diffraction ... so check your notes.