I o thin lenses with focal lengths of magnitude 18.0 cm the first diverging and the second converging. are placed LOO cm apart. An object 4 10 mm ta lis pa ed 4 75 m·to heeft ofthe st ive ging as (a) Where is the image formed by the first lens located? distance locationSelectof the first lens Cm (b) How far from the object is the final image formed? cm () Is the fnal image ealo virtual? real virtual (d) What is the height of the final image? cm Is the final image erect or inverted? erect O inverted
Solution
here,
a)
for the first lens
focal length , f1 = - 18 cm
object distance , do1 = 4.75 cm
let the image distance be di1
using the lens formula
1/f1 = 1/di1 + 1/do1
- 1/18 = 1/di1 + 1/4.75
solving for di1
di1 = - 3.76 cm
the image formed by first lens is 3.76 cm to the left of lens
b)
for the seccond lens
focal length , f2 = 18 cm
object distance , do2 = 11 - ( - 3.76) = 14.76 cm
let the image distance be di2
using the lens formula
1/f2 = 1/di2 + 1/do2
1/18 = 1/di2 + 1/14.76
di2 = - 82 cm
c)
the final image is virtual
d)
the height of image , hi = ho * ( di1/do1) * ( di2/do2)
hi = 4.1 * (- 3.76/4.75)*(-82/14.76)
hi = 18 cm
e)
the image is erect
.
Ecological Succession. ( ECOSYSTEM, B. Pharmacy, 1st Year, Sem-II, Environmen...
I o thin lenses with focal lengths of magnitude 18-0 cm the first dive.docx
1. I o thin lenses with focal lengths of magnitude 18.0 cm the first diverging and the second
converging. are placed LOO cm apart. An object 4 10 mm ta lis pa ed 4 75 m·to heeft ofthe st
ive ging as (a) Where is the image formed by the first lens located? distance locationSelectof the
first lens Cm (b) How far from the object is the final image formed? cm () Is the fnal image ealo
virtual? real virtual (d) What is the height of the final image? cm Is the final image erect or
inverted? erect O inverted
Solution
here,
a)
for the first lens
focal length , f1 = - 18 cm
object distance , do1 = 4.75 cm
let the image distance be di1
using the lens formula
1/f1 = 1/di1 + 1/do1
- 1/18 = 1/di1 + 1/4.75
solving for di1
di1 = - 3.76 cm
the image formed by first lens is 3.76 cm to the left of lens
b)
for the seccond lens
2. focal length , f2 = 18 cm
object distance , do2 = 11 - ( - 3.76) = 14.76 cm
let the image distance be di2
using the lens formula
1/f2 = 1/di2 + 1/do2
1/18 = 1/di2 + 1/14.76
di2 = - 82 cm
c)
the final image is virtual
d)
the height of image , hi = ho * ( di1/do1) * ( di2/do2)
hi = 4.1 * (- 3.76/4.75)*(-82/14.76)
hi = 18 cm
e)
the image is erect
