Question

Two systems are formed from a converging lens and a diverging lens, as shown in parts a and b of the drawing. (The point labeled "Fconverging" is the focal point of the converging lens.) An object is placed inside the focal point of lens 1 at a distance of 7.30 cm to the left of lens 1. The focal lengths of the converging and diverging lenses are 15.00 and -20.0 cm respectively. The distance between the lenses is 50.0 cm. Determine the final image distance for each system, measured with respect to lens 2.

Answer 1

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Two systems are formed from a converging lens and a diverging lens, as shown in partsaandbof the drawing.(The point labeled "F_converging" is the focal point of the converging lens.) An object is placed inside the focal point of lens 1at a distance of5.20 cmto the left of lens 1. The focal lengths of the converging and diverging lenses are 15.00 and-20.0cmrespectively. The distance between the lenses is50.0 cm. Determine the final image distance for each system, measured with respect tolens 2.

ANS:focal length of converging lens f1 = 15 cm
focal length of diverging lens f2 = -20 cm
distance between two lenses d = 50 cm
a) object distance of convergong lens p = 5.2 cm
Thin lens equation is
1/ P + 1 / Q = 1 / f1
1/ Q = 1 / 15 - 1 / 5.2
image distance of converging lens is
Q = -7.959 cm
object distance of diverging lens P ' = d - Q
= 50 + 7.959 = 57.959
Thin lens equation is
1 / P' + 1 / Q' = 1 / f2
1 / 57.959 + 1 / Q' = -1 /20
Q' = -14.86 cm
b)
object distance of diverging lens p = 5.2 cm
Thin lens equation is
1/ P + 1 / Q = -1 / 20
1/ Q = -1 / 20 - 1 / 5.2
image distance of converging lens is
Q = -40.12 cm
object distance of diverging lens P ' = d - Q
= 50 + 4.12= 54.12cm
Thin lens equation is
1 / P' + 1 / Q' = 1 / f2
1 / 54.12+ 1 / Q' = 1 /15
Q' = 20.75 cm