Question

A diverging lens has a focal length of magnitude 19.8 cm. (a) Locate the images for each of the following object distances 39.6 cm dstanice Your response daffers significandy from the correct answer: Reswork your soluation from the begining and check each step caruly,.om 19.8 cm location [n 9.9 ơn (b) Is the image for the object at distance 39.6 real or virtual? real e virtual is the image for the object at distance 19.8 real or virtual? O real e virtual Is the image for the object at distance 9.9 real or virtual? O real e virtual Is the image for the object at distance 39.6 upright or inverted? e upright o inverted Is the image for the object at distance 19.8 upright or inverted? upright O inverted ts the image for the object at distance 9.9 upright or inverted? upright D invertned d) Find the magnification for the object at distance 39.6 cm. Find the magnification for the object at distance 19.8 cm. Find the magnfication for the object a: dstance 9.9 cm.

Answer 1

Diverging lens focal length f = -19.8 cm [a] object distance = 39.6 cm from lens formula 1/f = 1/v - 1/u rightarrow 1/v - 1/f + 1/u V = uf/u + f v = (-39.6) times (-19.8)/-39.6 + (-19.8) = -13.2 cm v = -13.2 cm location in from of when object distance u = -19.8 cm then from v = u_f/u + f = (-19.8) times (-19,8)/-19.8 + (-19.8) = -9.9 cm v = -9.9 cm location in front of when object distance u = 9.9 cm v = -6.6 cm location in front of [b] when u = -39.6 cm then image is virtual when u = 19.8 cm the image is virtual when u = -9.9 cm then image is virtual \r\n [c] when object distance u = -39.6 cm then image is up right when u = -19.6 cm then image is up right when u = -9.9 cm then image is up right [d] when object distance