Question

an object is kept at a distance of 10cm from a convex mirror whose focal length is 15cm . find the position, nature and magnification​

Answer 1

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An object is placed at a distance of 10cm from a convex mirror of focal length 15cm. ... Thus, the position of image is at a distance of 6 cm from the convex mirror on its right side (behind the mirror). Since the image is formed behind the convex mirror, therefore, the nature of image is virtual and erect....~~~~

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Answer 2

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 .