40.

(a) The “object” for the mirror which results in that box-image is equally in front of the mirror (4 cm).

This object is actually the first image formedby the system (producedby the first transmission
through the lens); in those terms, it corresponds to i

1

= 10

− 4 = 6 cm. Thus, with f

1

= 2 cm,

Eq. 35-9 leads to

1

p

1

+

1

i

1

=

1

f

1

=

⇒ p

1

= 3.00 cm .

(b) The previously mentionedbox-image (4 cm behindthe mirror) serves as an “object” (at p

3

= 14 cm)

for the return trip of light through the lens (f

3

= f

1

= 2 cm). This time, Eq. 35-9 leads to

1

p

3

+

1

i

3

=

1

f

3

=

⇒ i

3

= 2.33 cm .