An object is 15cm left of a converging lens with focal length 10 cm. A diverging lens with focal length 20 cm is placed 20 cm right of the first lens. Find whether the final image is real or virtual and determine its position and magnification relative to the original object plz show working
Answers
Answer:
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Given,
Object distance, u=−20 cm
Focal length, f=16 cm
Using lens formula,
f
1
=
v
1
−
u
1
We have,
⇒
16
1
=
v
1
−
−20
1
⇒
v
1
=
16
1
−
20
1
⇒
v
1
=
20×16
20−16
=
320
4
⇒v=80 cm
Now, this image acts as object for the second lens.
Let the separation between the lens be d cm.
Object distance, u=80−d cm
Focal length, f=16 cm
⇒
16
1
=
v
1
−
80−d
1
⇒
v
1
=
16
1
+
80−d
1
⇒
v
1
=
16(80−d)
96−d
⇒v=
96−d
16(80−d)
We know,
m=
u
v
Magnification of first image, m
1
=
−20
80
=−4
Magnification of second image, m
2
=
80−d
(
96−d
16(80−d)
)
=
96−d
16
Given, image formed by the combination is of the same size and orientation as the original object.
⇒m
1
×m
2
=1
⇒−4×
96−d
16
=1
⇒−64=96−d
⇒d=96+64=160 cm
Hence, the correct answer is 160 cm
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