Question

7. A lens of focal length 15 cm produced an image of magnification +3 then state the position of object (u) and position of image (v) from the optic centre of a lens? ​

Answer 1

Explanation:

Here object distance u=−15 cm

v=30 cm

So, using the lens' formula

f

1

=

v

1

−

u

1

f

1

=

30

1

−

−15

1

f=10 cm

When the focal become half f

1

=f/2=5 cm

Let say image forms at v

1

So

f

1

1

=

v

1

1

−

u

1

5

1

=

v

1

1

−

−15

1

v

1

=7.5 cm

If the image has to form at the earlier position,

Let say object is placed at u

1

v=30 cm

f

1

1

=

v

1

−

u

1

1

5

1

=

30

1

−

u

1

1

u

1

=−6 cm

So initially it was at 15 cm and now it should be placed at 6 cm from lens.

So it need to move 15−6=9 cm towards lens.