An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the Image formed.
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Answer:
Given,
Height of object =5cm
Position of object, u=−25cm
Focal length of the lens, f=10cm
Position of image, v=?
We know that,
v
1
−
u
1
=
f
1
v
1
+
25
1
=
10
1
v
1
=
10
1
−
25
1
So,
v
1
=
50
(5−2)
That is,
v
1
=
50
3
So,
v=
3
50
=16.66cm
Thus, distance of image is 16.66cm on the opposite side of lens.
Now, magnification =
u
v
That is,
m=
−25
16.66
=−0.66
Also,
m=
heightofobject
heightofimage
or
−0.66=
5cm
heightofimage
Therefore, Height of image =3.3cm
The negative sign of the height of image shows that an inverted image is formed.
Thus, position of image is at 16.66cm on opposite side of lens.
Size of image =−3.3cm at the opposite side of lens
Nature of image is real and inverted.
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