An object of height 2cm is placed in front of a convex lens of focal length 20cm at a distance of 15cm from it. Find the position and magnification of the image.
Answers
Answer:
Explanation:
Given:-
Height of the object = 2cm
Focal length = 20cm
Object distance = 15cm
Type of lens used = Convex
To Find:-
The Magification and Nature of the image.
Solution:-
For a convex lens:-
• Focal length (f) = +ve = 20cm
• Object distance = -ve = -15cm
Using lens formula:-
\boxed{ \sf \leadsto \frac{1}{f} = \frac{1}{v} - \frac{1}{u} }
⇝
f
1
=
v
1
−
u
1
\sf \implies \dfrac{1}{20} = \dfrac{1}{v} - \dfrac{1}{ - 15}⟹
20
1
=
v
1
−
−15
1
\sf \implies \dfrac{1}{20} = \dfrac{1}{v} +\dfrac{1}{ 15}⟹
20
1
=
v
1
+
15
1
\sf \implies \dfrac{1}{v} = \dfrac{1}{20} - \dfrac{1}{ 15}⟹
v
1
=
20
1
−
15
1
\sf \implies \dfrac{1}{v} = \dfrac{3 - 4}{60}⟹
v
1
=
60
3−4
\sf \implies \dfrac{1}{v} = \dfrac{-1}{60}⟹
v
1
=
60
−1
\sf \implies v = -60cm⟹v=−60cm
\sf \underline{\:\:\underline{\: Image \: distance \: (v) = -60cm\:}\:\:}
Imagedistance(v)=−60cm
\sf Magnification = \dfrac{v}{u}Magnification=
u
v
\sf \implies m = \dfrac{60}{-15}⟹m=
−15
60
\sf \implies m = 4⟹m=4
Since, the magnification is positive
Nature of the image is:-
Virtual
Erect
Enlarge
There is your answer
Hope it helps you
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