Question

Two-lens systems. In the figure, stick figure O (the object) stands on the common central axis of two thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closer to O, which is at object distance p₁. Lens 2 is mounted within the farther boxed region, at distance d. For this problem, p₁ = 10 cm, lens 1 is diverging, d = 16 cm, and lens 2 is converging. The distance between the lens and either focal point is 5.4 cm for lens 1 and 6.9 cm for lens 2. (You need to provide the proper sign).
Find (a) the image distance i₂ for the image produced by lens 2 (the final image produced by the system) and (b) the overall lateral magnification M for the system, including signs. Also, determine whether the final image is (c) real or virtual, (d) inverted from object O or noninverted, and (e) on the same side of lens 2 as object O or on the opposite side.

Answer 1

for the lens 1 we have object distance p₁ and image distance i₁ and focal length f₁

1/ f₁ = 1/ i₁ + 1/ p₁

we ignore the lens 1 but treat the image formed by lens 1 ignore the lens 1 as th object for lens 2

here distance p₂ is the distance between lens 2 and location of first image, here distance for lens 2 image is i_2.

here we have a equation

1/ f₂ = 1/ i₂ + 1/ p₂

here lens 1 is diverging lens, so focal length should be always negetive , f₁ = - 5.4 cm ,

i₁ = (p_{1 *} f₁ ) / (p₁ - f₁ )

= ( 10 * (-5.4) ) / ( 10- (-10))

= -54 / 20

i_{1 = - 2.7 cm}

here image distance of lens 1 is negetive , it means is on the same side of length , here image 1 is object for lens 2 , object distance for lens 2 ,

p₂ = d - i₁

p₂  = 16 - (- 2.7 )

  p₂  = 18.7 cm

here image distance for the lens 2 ,

i₂ =  (p_{2 *} f₂) / (p₂ - f₂ )

here second lens is converging lens, so focal length should be always positive. s f₂ = 6.9 cm

     i₂ =  (p_{2 *} f₂) / (p₂ - f₂ )

= ( 18.7 * 6.9 ) / (18.7 - 6.9 )

= 129.03 / 11.8

i₂ = 10.93 cm

(b)

here we want to find the overall magnification , M

M= i₁ i₂ / p₁ p₂

= (-2.7 ) 10.93 / 18.7 * 10

= - 29.511 / 187

= - 0.157 cm

(c)

here the final image distance i₂ is positive so the image is real.

(d)  here the magnification value is negetive so the image is inverted .

(e) here the lens 2 is converging lens and object distance for lens 2 is positive. then the object 2 is on same side of object.