Consider a two lens system that consists of a converging lens (lens 1) with foca

Question

Consider a two lens system that consists of a converging lens (lens 1) with focal length f_1 = 5 cm, without proper sign, and a diverging lens (lens 2) of focal length f_2 = 3.07 cm, without proper sign. Lens 2 is located at 13.5 cm to the right of lens 1. D) Ignore lens 1, find the distance i_2 at which the image due to lens 2, I_2, is formed. E) Calculate the lateral magnification due to the compound system, M = m_1 m_2 M = ___ (with proper sign) Image I_2 is ___ with respect to the object O_1 INVERTED UPRIGHT Image I_2 is ___than object I_1 SMALLER LARGER

Explanation / Answer

D)

for diverging lens :

do = object distance = 12.14 + 13.5 = 25.64 cm

di = image distance = I2

f2 = focal length = - 3.07 cm

using the lens equation

1/do + 1/di = 1/f2

1/25.64 + 1/I2 = 1/(- 3.07)

I2 = - 2.74 cm

hence image is formed at distance 2.74 cm from lens 2

E)

for lens 1 :

for diverging lens :

do = object distance = 12.14 cm

di = image distance = I1

f1 = focal length = 5 cm

using the lens equation

1/do + 1/di = 1/f1

1/12.14 + 1/I1 = 1/(5)

I1 = 8.5 cm

m1 = magnification by lens 1 = - di/do = - 8.5/12.14 = - 0.7

force lens 2 :

do = object distance = 13.5 - 8.5 = 5 cm

di = image distance = I2

f2 = focal length = - 3.07 cm

using the lens equation

1/do + 1/di = 1/f2

1/5 + 1/I2 = 1/(- 3.07)

I2 = - 1.9 cm

m2 = magnification by lens 2 = - di/do = - (- 1.9)/5 = 0.38

lateral magnification by combination is given as

m = m1 m2 = (- 0.7) (0.38) = - 0.27

M = - 0.27

inverted : negative sign indicates inverted image

smaller : since M<1 , hence image is smaller

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