Question

11. A converging lens has focal length 2 po
25cm. An image of a pencil kept in
front of the lens is formed at a
distance of 50cm on a screen. the
magnification is:
O +2
02
O +1
O-1​

Answer 1

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