Question

The image of an object placed at 60 cm in front of a lens is obtained on a screen at a distance of 120 cm from it. Find the focal length of lens. What would be the height of the image if object is 5 cm high

Answer 1

Object distance u = -60 cm
Image distance v = +120 cm
Using lens formula, 1/f = 1/v - 1/u
1/f = 1/120 - ( -1/60 ) = 1/120 + 1/60 = 3/120 = 40 
f = +40 cm
H(image) / H(object) = v/u
Hi = -120/60 X 5 = -10 cm :)

Answer 2

Hi,

Object distance=u= -60cm

(It has minus sign because the object is placed in front of the lens like this:

So according to Cartesian (graph) rule the object is in 2nd quadrant with minus sign)

Image distance=120 (It has a positive sign as the image lies behind the lens which also means that it lies in 1st quadrant)

Now, use lens formula:

1/f=1/v-1/u

Substitute v and u

1/f=1/120-1/-60

After taking LCM and calcuting we get.....

1/f=1/4 ....... don't panic, just inverse the values on both side

f=4

Therefore, Focal length is 4cm

Hight of the image? Hmmmmm...use Magnification formula!!

m=h'/h=v/u

Where h'=height of the image

h=height of the object

Now substitute the values

h'/5=120/-60

h'=600/-10

We get...

h'= -10