Question

1

(a) How far from a 54.2 mm focal-length lens must an object be placed if its image is to be magnified 2.36? and be real?
_____ mm
(b) What if the image is to be virtual and magnified 2.36??
__________ mm

2

A diverging lens with a focal length of -14 cm is placed 14 cm to the right of a converging lens with a focal length of 18 cm. An object is placed 31 cm to the left of the converging lens.

(a) Where will the final image be located?
__________cm left of the converging lens
(b) Where will the image be if the diverging lens is 34 cm from the converging lens?
____________cm right of the diverging lens

Answer 1

1/f = 1/o + 1/i

(a)
i/o = 2.36
i = 2.36o

1/54.2 = 1/o + 1/(2.36o)
1/54.2 = 2.36/(2.36o) + 1/(2.36o)
1/54.2 = (2.36 + 1)/(2.36o)
o = 54.2*(2.36 + 1)/2.36
o = 77.16 mm

(b)
i/o = -2.36
i = -2.36o

1/54.2 = 1/o - 1/(2.36o)
1/54.2 = 2.36/(2.36o) - 1/(2.36o)
1/54.2 = (2.36 - 1)/(2.36o)
o = 54.2*(2.36 - 1)/2.36
o = 31.23 mm

2.

First for the converging lens:

1/s + 1/s'= 1/f
1/31 + 1/ s'= 1/18 ==> s'= 42.92cm

Thus the object distance for the second lens will be -(42.92 - 14) = -28.92cm. Hence using the thin lens formula for the second lens:

1/ (-28.92) + 1/s' = -1/14 ==> s'= -27.136cm.

The negative sign means that the image is 27.14 cm to the left of the second lens.

Second part: assuming the diverging lens is 34 cm from converging lens, the calculation for the diverging lens will be:

1/(-8.92) + 1/s' = -1/14 ==> s'= 24.58cm. The image will be 24.58 cm to the right of the diverging lens.