Question

A converging lens L_1 has a focal length of 10 cm. A 5cm tall object is located 15cm to the left of L_1. Construct a ray diagram indicating the position and vertical orientation of the image produced by lens L_1. Calculate this position and see that it matches your diagram. A diverging lens L_2 with a focal length of -20 cm is placed 40cm on the right of lens L_1. The image in (a) above now serves as the object for lens L_2 Construct a ray diagram to find the image (final image) produced by lens L_2. Calculate its position and see that it matches your diagram. Is this final image real or virtual? Can it be shown on a screen? Move the diverging lens L_2 to a new location which is 20 cm to the right of lens L_1. In the space below, sketch the two lenses, the original object, and the image produced by lens L_1. Calculate the location of the final image. (Ray diagram is not required). Is this final image real or virtual? Can it be shown on a screen?

Answer 1

a) Ray diagram Lens equation-=-+ V tu Lu f + u 10x-15 -15 10 30 cm Final image is real. It can be shown on a screen.

b) Ray diagram Image -10 cm Object tu Lens equation v= +1 -10x-20 -10-20 6.6 cm Final image is virtual. It cannot be shown on a screen.

C) 1 V u v20 10 200 10 v =-20 cm The image obtained is virtual and thus it cannot be shown on the screen