heya✌___answer this___An object 10cm long is placed at 15 cm from a convex lens of focal length 10cm.Find position and size of image.
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Answer:
position of a image is 6cm behind th
e convex mirror at a size of four.
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position of a image is 6cm behind th
\frac{1}{f} =
f
1
=
e convex mirror at a size of four.
hi = 10cmhi=10cm
u = - 15cmu=−15cm
f = 10cmf=10cm
v =v=
position \: of \: image =positionofimage=
size \: of \: image =sizeofimage=
by \: the \: mirror \: formula =bythemirrorformula=
\frac{1}{f} = \frac{1}{u} + \frac{1}{v}
f
1
=
u
1
+
v
1
\frac{1}{10} = \frac{1}{v} + \frac{1}{ - 15}
10
1
=
v
1
+
−15
1
\frac{1}{v} = \frac{1}{10} + \frac{1}{15}
v
1
=
10
1
+
15
1
\frac{1}{v} = \frac{2 + 3}{30}
v
1
=
30
2+3
\frac{1}{v} = \frac{5}{30}
v
1
=
30
5
\frac{1}{v} = \frac{1}{6}
v
1
=
6
1
v = 6v=6
\begin{gathered}so \: the \: position \: of \: the \: image \\ is \: 6cm \: behind \: the \: mirror \: convex \\ \: mirror.\end{gathered}
sothepositionoftheimage
is6cmbehindthemirrorconvex
mirror.
\frac{hi}{ho} = \frac{ - v}{u}
ho
hi
=
u
−v
\begin{gathered} \frac{hi}{10} = \frac{ - 6}{ - 15} = \frac{2}{5} \\ \frac{2 \times 10}{5} = \frac{20}{5} \end{gathered}
10
hi
=
−15
−6
=
5
2
5
2×10
=
5
20
hi = \frac{20}{5}hi=
5
20
the \: size \: of \: the \: image \: is \: 4cmthesizeoftheimageis4cm
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