# Physics 3: Images, Interference and Difraction

So three problems im 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 im lost is in the r1 and r2. would r2 be -r1 so we get 2r?

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 im 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 theres no limit?

http://ift.tt/1qW3EIK

## Leave a comment