1.

An object 5 cm in length is held 30 cm away from a converging lens of focal length20 cm. Draw the ray diagram and find the position, size and the nature of the image formed.​

Answer»

Answer:

Given,

Height of object =5cm

POSITION of object, u=−25cm

Focal length of the lens, f=10cm

Position of image, V=?

We know that,

v

1

−

u

1

=

f

1

v

1

+

25

1

=

10

1

v

1

=

10

1

−

25

1

So,

v

1

=

50

(5−2)

That is,

v

1

=

50

3

So,

v=

3

50

=16.66cm

Thus, distance of image is 16.66cm on the OPPOSITE side of lens.

Now, magnification =

u

v

That is,

m=

−25

16.66

=−0.66

Also,

m=

heightofobject

heightofimage

or

−0.66=

5cm

heightofimage

Therefore, Height of image =3.3cm

The negative sign of the height of image shows that an inverted image is formed.

Thus, position of image is at 16.66cm on opposite side of lens.

Size of image =−3.3cm at the opposite side of lens

Nature of image is real and inverted.


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