Question

A 2.0 cm tall object is placed perpendicular to the principal axis of convex lens of focal length 10 cm, then the nature of the image is​

Answer 1

According to the question:

Image distance, v=−10 cm

Focal length, f=−15 cm (since, concave lens)

Let the object distance be u.

By lens formula:

v

1

−

u

1

=

f

1

[4pt]

⇒

−10 cm

1

−

u

1

=

−15 cm

1

[4pt]

⇒

u

1

=−

−15 cm

1

−

10 cm

1

[4pt]

⇒

u

1

=

30 cm

2−3

=

30 cm

−1

[4pt]

⇒

u

1

=

30 cm

−1

∴u=−30 cm.

Thus, object should be placed at 30 cm in front of lens, in the same side as image.

Now,

Height of object, h

1

=2 cm

Magnification, m=

h

1

h

2

=

u

v

Putting value of v and u:

Magnification, m=

2 cm

h

2

=

−30 cm

−10 cm

⇒

2 cm

h

2

=

3

1

⇒h

2

=

3

2

cm=0.67 cm

Height of image is 0.67 cm

Thus, the image is virtual, diminished, and erect and one-third the size of object.

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