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**

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