1 answer:
Explanation:
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,
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
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Answer:
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If it starts from 0m/s...
s=?
u=0
a=-10
t=8
s=ut +1/2at^2
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