Answer:
a) I2 = 3 (o-10) / (o- 30)
, b) h ’/h= 3 (o-10) / o (o-30)
Explanation:
The builder's equation is
1 / f = 1 / o + 1 / i
Where f is the focal length, or e i are the distance to the object and image, respectively
As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm
Let's calculate the position of the image of this lens
1 / i1 = 1 / f - 1 / o
1 / i1 = 1/15 - 1 / o
i1 = o 15 / (o-15)
Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second
o2 = d-i1
We write the builder equation
1 / f2 = 1 / o2 + 1 / i2
1 / i2 = 1 / f2 -1 / o2
1 / i2 = 1 / f2 - 1 / (d-i1)
1 / i2 = 1/20 - 1 / (d-i1) (1)
Let's evaluate the last term
d-i1 = d - 15 o / (o-15)
d-i1 = (d (o-15) - 15 o) / (o-15)
d- i1 = (30 or -30 15 -15 o) / (o-15)
d-i1 = (15 or - 450) / (o- 15)
d-i1 = = (15 or -450) / (o-15)
replace in 1
1 / i2 = 1/20 - (or - 15) / (15 or -450)
1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)
1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)
1 / i2 = (-5 or -150) / (15 or -150)
1 / i2 = (or -30) / (3 or - 30)
I2 = 3 (o-10) / (o- 30)
Part B
The height of the image, we use the magnification equation
m = h ’/ h = - i / o
h ’= - h i / o
In our case
h ’= h i2 / o
h ’= h 3 (o-10) / o (o-30)
If they give the distance to the object it is easier
and the frequency f: