Question

For each scenario, draw the ray diagram (roughly to scale) and show your math, including original equations used. Label all values, including: s, s' , h, h' , f.

• Using a 5 cm tall object and a +10 cm lens, create a real image that is half the height of the object.

• Using a 5 cm tall object and a +10 cm lens, create a virtual image that is twice the height of the object.

Answer 1

for first part :

My sign convention is cartesian convention, where the lens is at origin and measurement is positive in the direction of incident light.

label : s = object distance

s' = image distance

h = height of object

h' = height of image

f = focal length

given :

h = 5 cm

f = 10 cm

M = magnification = - 1/2 [since the image formed is real of a real object]

Now, as we know M = h'/h = s'/s = (f - s')/f

So, puting the value of M = - 1/2 and taking "s" to be negative,

we can say, s' = s/2

also, M = (f - s')/f

=> - 1/2 = (10 - s')/10

=> s' = 15 cm.

also, s = 2 s' = 30 cm. [ just the value of object distance]

and h' = - h/2 = - 2.5 cm [ -ve sign indicates, the image formed is inverted].

2nd Part:

label : s = object distance

s' = image distance

h = height of object

h' = height of image

f = focal length

given :

h = 5 cm

f = 10 cm

M = magnification = + 2 [since the image formed is virtual of a real object]

Now, as we know M = h'/h = s'/s = (f - s')/f

So, puting the value of M = + 2 and taking "s" to be negative, ( s' will also be negative, which is calculated later)

we can say, s' = 2 s

also, M = (f - s')/f

=> +2 = (10 - s')/10

=> s' = - 10 cm.

also, s = s'/2 = 5 cm. [just the value of object distance]

and h' = 2h = +10 cm [ +ve sign indicates, the image formed is erect or upright].