Question

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Answer 1

Answer:

Here, focal length of concave lens,

f=−15cm

f=-15cm

.

object distance,

u=?

u=?

, image distance,

v=−10cm

v=-10cm

, magnification of lens,

m=?

m=?

As

1

f

=

1

v

−

1

u

,

1

u

=

1

v

−

1

f

=

1

−10

+

1

15

=

−1

30

or

u=−30cm

1f=1v-1u,1u=1v-1f=1-10+115=-130oru=-30cm

Thus the object should be placed at a distance of 30 cm on the left side of the concave lens.

Linear magnification,

m=

v

u

=

− 10

− 30

=

1

3

m=vu=-10-30=13

.

The positive sign of m shows that the image is virtual and erect, and its size is

(1/3)

(1/3)

of the size of the object