Question

An object 4.1 mm high is on the optical axis of two lenses with focal lengths f₁ = +5.0 cm, f₂ = +10 cm. The object is 5.9 cm to the left of the first lens, and the second lens is 33 cm to the right of the first lens. Find the position (relative to the second lens) and the size of the final image by calculation.

position	-0.23 cm
size

Find the position (relative to the second lens) and the size of the final image by the graphical method.

Answer 1

v= image distance , u= object distance

1/v -1/u = 1/f

1/v- 1/ -5.9 = 1/5

from here first image position v≈32.7778 cm in righ side of Lens1

magnification =m1 = v/u = 32.7778/ -5.9 = -5.55555932

now this image act like object for 2nd lens so distance of this image from 2nd lens

33-32.7778 = 0.2222 cm

for second image  

1/v-1/u = 1/f

1/v- 1/-0.2222 = 1/10

v≈-0.227249 cm in left so 2nd lens

position (relative to the second lens) = 0.227249 cm in left answer

magnification m2 = v/u = -0.227249 /-0.2222 =1.02272277

total magnification M = m1*m2 = -5.55555932*1.02272277 = -5.68179702

height of image = M*4.1 mm = -5.68179702*4.1 = 23.2953678 mm

height of image = 23.2953678 mm   answer

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