Question

The figure shows a combination of two lenses.(Figure 1) 

Part A 

Find the position of the final image of the 1.0-cm-tall object.
Express your answer in centimeters to two significant figures. 

Part B 

Find the size of the final image of the 1.0-cm-tall object.
Express your answer in centimeters to two significant figures.

Part C Find the orientation of the final image of the 1.0-cm-tall object. 

• erect 

• inverted

Answer 1

Date using lens formula for convexe lens. f- focal length V- Distance of u- " " image object - Vi= -20cm Image formed by conven lens will act as - Virtual o object for a concave lens ( U = Nit12 Again using lens formula fol concave. -8 V (96+12) - -/+ 1 8 32 1.= -10.7 cm (12-10.7) + 4 cm from 5.3 cm teft of object Right - (In between both lenses over all magnification (m) т - mix м2. m2 = √ -10.7 = 0.33 - U2 -32 m = 5x0.33 m = 1667 size of image is 1067 cm ere overall magnification is positive Image is erect. erec it means L1.11

Answer 2

a) image distance from object= 9.6 cm

b) h'= 1 cm

Answer 3

apply for converging lens 1/f = 1/u1 + 1/v1

1/v1 = 1/5- 1/4

v1 = -20 cm (image distance for the first lense)

f2 = - 8 cm

u2 = -20-12 = -32 cm (object distance for the second lense)

1/u2 + 1/v2 = 1/f2

1/v2 = 1/f2 - 1/u2

1/v2 = -1/8 - 1/32

v2 = -6.4 cm

from object 12-6.4 +4    9.6 cm

B) magnification, m = (-v1/u1)*(-v2/u2)

= (20/4)*(6.4/32)

= 1

imahe height = m*object height

= 1*1

= 1 cm

C) erect