I posted this question earlier, but unfortuantely the answers for the distances
ID: 1486820 • Letter: I
Question
I posted this question earlier, but unfortuantely the answers for the distances were incorrect. Please help. Thanks!
Find the image distance and the magnification for a thin lens in air when the object distance is -23 cm and the lens is a converging lens that has a focal length of 19 cm. Describe the image - is it virtual or real, upright or inverted?
realvirtual
(c) Repeat Part (b) if the object distance is, instead, -9 cm and the lens is diverging and has a focal length (magnitude) of 30 cm.
realvirtual
Explanation / Answer
(a).Object distance u = -23 cm
Focal length f = 19 cm
Image distance v = ?
from the relation ( 1/v) -(1/u) = (1/f)
(1/v) = (1/f) +( 1/ u)
= (1/19) +(1/-23)
=(1/19) -(1/23)
= (23-19) / (19 x23)
So, v = (19 x23) / (23-19)
= 109.25 cm
The image is real and inverted
(b).object distance u = -9 cm
Focal length f = -30 cm
From the relation
( 1/v) +(1/u) = (1/f)
(1/v) = (1/f) -( 1/ u)
= (1/-30) -(1/-9)
=(1/-30) +(1/9)
= -0.033333 +0.111111
= 0.077777
v = 12.85 cm
The image is real and virtual