an object is kept at a distance of 10cm from a convex mirror whose focal length is 15cm . find the position, nature and magnification
Answers
Given that
u= -10 cm
f= 15 cm
We need to find v
We know the formula
1/f = 1/v + 1/u
Therefore, 1/v= 1/f – 1/u
Substituting the known values in the above equation we get,
1/v= 1/15 + 1/10
1/v = (2+3)/30 (lcm)
Hencev=5/30 = 6 cm
v= 6cm
Nature of the image: The image is formed behind the mirror and it is virtual and erect.
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Answer:
Answer :-
(•) Image distance = 30 cm
(•) Nature is virtual and erect.
(•) magnification = 0.6
To Find :-
→ Position of image ,nature and magnification.
Explanation :-
Given that ,
Focal length (f) = 15 cm
object distance (u) = -10 cm ( left )
image distance (v) = ?
magnification (m) = ?
Apply mirror formula
\begin{gathered} \implies \boxed{ \sf{ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} }} \\ \therefore \: \\ \\ \implies \sf{ \frac{1}{ 15} = \frac{1}{v} + \frac{1}{ - 10} } \\ \\ \implies \sf{\frac{1}{v} = \frac{1}{10} + \frac{1}{15} } \\ \\ \implies \sf{ \frac{1}{v} = \frac{3 + 2}{30} } \\ \\ \implies \sf{ \frac{1}{v} = \frac{5}{30} } \\ \\ \implies \boxed{ \sf{v(image \: distance) = 6 \: cm \: }}\end{gathered}
⟹
f
1
=
v
1
+
u
1
∴
⟹
15
1
=
v
1
+
−10
1
⟹
v
1
=
10
1
+
15
1
⟹
v
1
=
30
3+2
⟹
v
1
=
30
5
⟹
v(imagedistance)=6cm
Nature :-
→ Because v is positive hence image is erect and virtual .
Magnification :-
We know that ;
\begin{gathered} \to \sf{magnification \: = \frac{ - v}{u} = \frac{ - 6}{-10} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 0.6 \: \end{gathered}
→magnification=
u
−v
=
−10
−6
=0.6
We know that magnification for a convex mirror is less then 1 .