Question 10
(a) Find the position and size of the
virtual image formed when an object
2 cm tall is placed 20 cm from:
(1) a diverging lens of focal length 40
cm.
(ii) a converging lens of focal length
40 cm.
(b) Draw labelled ray diagrams to
show the formation of images in
cases (i) and (ii)above (The diagrams
may not be according to scale).
Object
f
Image
Object Inside
local point
Formation of gee.
Conves
ons
ago
Virtual
Answers
Answer:
Height of object = 2 cm
Object distance = 20 cm
Focal length of diverging lens = -40 cm
Focal length of converging lens = 40 cm
(I). For converging lens,
We need to calculate the image distance
Using formula of lens
\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}
f
1
=
v
1
−
u
1
Put the value into the formula
\dfrac{1}{40}=\dfrac{1}{v}-\dfrac{1}{-20}
40
1
=
v
1
−
−20
1
\dfrac{1}{v}=\dfrac{1}{40}-\dfrac{1}{20}
v
1
=
40
1
−
20
1
\dfrac{1}{v}=-\dfrac{1}{40}
v
1
=−
40
1
v=-40\ cmv=−40 cm
Negative sign shows that the image formed is virtual and erect.
We need to calculate the height of image
Using formula of magnification
m=\dfrac{v}{u}=\dfrac{h'}{h}m=
u
v
=
h
h
′
Where, v = image distance
u = object distance
h' = height of image
h = height of object
Put the value into the formula
\dfrac{-40}{-20}=\dfrac{h'}{2}
−20
−40
=
2
h
′
h'=\dfrac{40\times2}{20}h
′
=
20
40×2
h'=4\ cmh
′
=4 cm
The size of image is 4 cm.
(II). For converging lens
We need to calculate the image distance
Using formula of lens
\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}
f
1
=
v
1
−
u
1
Put the value into the formula
\dfrac{1}{-40}=\dfrac{1}{v}-\dfrac{1}{-20}
−40
1
=
v
1
−
−20
1
\dfrac{1}{v}=\dfrac{1}{-40}-\dfrac{1}{20}
v
1
=
−40
1
−
20
1
\dfrac{1}{v}=-\dfrac{3}{40}
v
1
=−
40
3
v=-\dfrac{40}{3}\ cmv=−
3
40
cm
v=-13.33\ cmv=−13.33 cm
Negative sign shows that the image formed is virtual and erect.
We need to calculate the height of image
Using formula of magnification
m=\dfrac{v}{u}=\dfrac{h'}{h}m=
u
v
=
h
h
′
Where, v = image distance
u = object distance
h' = height of image
h = height of object
Put the value into the formula
\dfrac{-40}{-20\times3}=\dfrac{h'}{2}
−20×3
−40
=
2
h
′
h'=\dfrac{40\times2}{20\times 3}h
′
=
20×3
40×2
h'=1.33\ cmh
′
=1.33 cm
The size of image is 1.33 cm.
Hence, (I). The position of image is formed 40 cm distance from the converging lens in left side.
The size of image is 4 cm
(II). The position of image is formed -13.3 cm distance from the diverging lens in left side.
The size of image is 1.33 cm.
Explanation:
hope it helps
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