Question

an object 5cm in length is placed 25 cm away from a converging lens of focal length 10 cm. find the position size and nature of the image formed.

Answer 1

ho=5cm
u=-25cm
f=10cm
by lens formula,
1/v-1/u=1/f
1/v=1/f+1/u
=1/10+1/(-25)
=(5-1)/50=4/50=2/25
v=25/4=6.25cm

hi/ho=v/u
hi/5=6.25/(-25)
=-1.25
hi=-1.25

image is inverted and real.

Answer 2

Given, height of object = 5cm

Position of object, u = - 25cm

Focal length of the lens, f = 10 cm

Hence, position of image, v =?

We know that,

1/v - 1/u = 1/f

1/v + 1/25 = 1/10

So, 1/v = 1/10 - 1/25

S0, 1/v = (5 - 2)/50

That is, 1/v = 3/50

So, v= 50/3 = 16.66 cm

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

Now, magnification = v/u

That is, m = 16.66/-25 = -0.66

Also, m= height of image/height of object

Or, -0.66 = height of image / 5 cm

Therefore, height of image = -3.3 cm

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

Thus, position of image = At 16.66 cm on opposite side of lens

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