A diverging lens (f = -11.5 cm) is located 20.0 cm to the left of a converging lens (f = 33.5 cm). A 3.00 cm tall object stands to the left of the diverging lens, exactly at its focal point.
(a) Determine the distance of the final image relative to the converging lens (in cm)
(b) What is the height of the final image (including proper algebraic sign)?
2007-03-06 06:06:41 · 1 answers · asked by christian m 2 in Science & Mathematics ➔ Physics
a) we will first search the image of the object which was -11.5 cm bat the left of the diverging lens . Convention when an object is at the left the sign is negative. Lets p distance of object to divergin lens in m p =-0.115
the vergence or power of the diverging lens is V= -1/0.115 =-8.7D
you have 1/p' = -1/0.115-1/0.115= -8.7-8.7 =-17.4D and p'=-0.058m
The image of the 3cm object has a size i/0= p'/p = 3*0.058/0.115=1.5cm and that image will be the objet for converging lens
Now, you must find for the second lens the image of an object which is 20+5.8=25.8cm =0.258m before and a size of 1.5 cm
1/p' = 1/p +V p = -0.258m V= 1/0.335= -3D
1/p' = -1/0.258+3 = 0.89 and p' =-1.12m
the image is at the left of the converging lens at 1.12m
the size i/o = p'/p = -1.12/-0.258= 4.34
last you multiply 4.34* 1.5 =6.51 cm
2007-03-06 06:55:00 · answer #1 · answered by maussy 7 · 0⤊ 0⤋