Question

A 5.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10cm. The distance of the object from the lens is 25 cm. Find the nature, position & size of the image. Also find its magnification​

Answer 1

Explanation:

According to the question:

Object distance, u=−12 cm

Focal length, f=18 cm (since, convex lens)

Let the Image distance be v.

By lens formula:

v

1

−

u

1

=

f

1

[4pt]

⇒

v

1

−

−12 cm

1

=

18 cm

1

[4pt]

⇒

v

1

=

−12 cm

1

+

18 cm

1

[4pt]

⇒

v

1

=

36 cm

−3+2

[4pt]

⇒

v

1

=−

36 cm

1

[4pt]

∴v=−36 cm

Thus, image is obtained at 36 cm on the same side of the mirror as the object.

Now,

Height of object, h

1

=5 cm

Magnification, m=

h

1

h

2

=

u

v

Putting values of v and u:

Magnification m=

5 cm

h

2

=

−12 cm

−36 cm

⇒

5 cm

h

2

=3

[4pt];

⇒h

2

=3×5 cm=15 cm

Thus, the height of the image is 15 cm and the positive sign means the image is virtual and erect.

Thus virtual, erect and enlarged image is formed.

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