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A object is located 16.0 cm in front of a converging lens whose focal length is 12.0 cm. A diverging lers ength of 10.0 cm is located 20.0 cm to the right of the converging lens. 26. An object i with a focal l a) Find the location of the final image relative to the diverging lens. b) Specify the characteristics of the final image: Numeric value of the total magnification of the two lenses. Type of final image, real or virtual. Orientation of final image, upright or inverted.

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Answer:

first lens v = 48 cm

second lens v = -15.6 cm

magnification =  1.67

final image is virtual

and final image is upright

Explanation:

given data

distance = 16 cm

focal length f1 = 12 cm

focal length f2 = 10.0 cm

to find out

location of the final image and magnification and Type of image

solution

we apply here lens formula that is

1/f = 1/v + 1/u     .....................1

put here all value and find v  for 1st lens

1/12 = 1/v + 1/16

v = 48 cm

and  find v  for 2nd lens

here u = 20- 48 = -28

- 1/10 = 1/v - 1/28

v = -15.6 cm

and

magnification = first lens (v/u) ×  second lens ( v/u)

magnification =  (-15.6/-28) ×  ( 48/16)

magnification =  1.67

so here final image is virtual

and final image is upright

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