Question

An object is 5cm in length
is held 25cm
from courying lens
at
focal length
10 cm draw
the ray diagrom
and find the position size and
nature of image
​

Answer 1

Given that

The height of object = 5cm

Position of object, u = – 25cm

Focal length of the lens, f = 10 cm

We need to find

The position of image, v =?

Size of the image

Nature of the image

Formula

We know that

1/v – 1/u = 1/f

Substituting the known values in the above equation we get,

1/v + 1/25 = 1/10

=> 1/v = 1/10 – 1/25

=> 1/v = (5 – 2)/50

Hence, 1/v = 3/50

So, v= 50/3 = 16.66 cm

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

Now, we know that

Magnification = v/u

Hence, m = 16.66/-25 = -0.66

Also, we know that

m= height of image/height of object

Or, -0.66 = height of image / 5 cm

Hence, height of image = -3.3 cm

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

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

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

Nature of image – Real and inverted