Question

Two thin lenses with a focal length of magnitude 11.0 cm, the first diverging and the second converging, are located 8.25 cm apart. An object 1.60 mm tall is placed 18.3 cm to the left of the first (diverging) lens. Part A How far from this first lens is the final image formed? Part B Is the final image real or virtual? Part C What is the height of the final image? Part D Is it upright or inverted?

Answer 1

Given,

f1 = -11 cm ; f2 = 11 cm ; d = 8.25 cm ; h = 1.6 mm ; o1 = 18.3 cm

A)We know from lens eqn

1/f = 1/i + 1/o

i = o x f/(o - f)

i1 = 18.3 x -11 /(18.3 + 11) = -6.87 cm

o2 = 8.25 - (-6.87) = 15.12 cm

i2 = 15.12 x 11 /(15.12 - 11) = 40.37 cm

D = 40.37 + 8.25 = 48.62 cm

Hence, D = 48.62 cm

b)Real

c)we know that

M = -i/o

m1 = -(-6.87)/18.3 = 0.38

m2 = -40.37/15.12 = -2.67

M = m1 x m2

M = 0.38 x -2.67 = -1.015

h' = M h = 1.015 x 1.6 = 1.623 cm

Hence, h' = 1.623 cm

d)Inverted