(a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all? (b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.
Answers
Given :
Focal length of the convex lens= f1 = 30 cm
Focal length of the concave lens,=f2 = −20 cm
Distance between the two lenses=d = 8.0 cm
(a) When the parallel beam of light is incident on the convex lens first:
According to the lens formula: 1/f1=1/v1-/1u1
Where
u1= object distance = ∞
f1=30cm
v1= image distance
1/v1= 1/f1+1/u1
=1/30 -1/∞
1/v1=1/30
V1= 30cm
the image will act as a virtual object for the concave lens.
Applying lens formula to the concave lens,
1/f2=1/v2-1/u2
Where
Object distance= 30-d=30-8=22cm
V2= image distance
1/v2=1/22-1/20
=10-11/220
=-1/220
V2=- 220cm
The parallel incident beam appears to diverge from a point that is = (220-d/2)= (220 - 8/2) = 220-4 = 216 cm from the centre of the combination of the two lenses.
(b)
According to lens formula
1/v2 - 1/u2 = 1/f2
where, u2 = Object distance = - ∞,
v2 = Image distance
1/v2 = 1/f2 + 1/u2
= 1/-20 +1/- ∞ = -1/20
v2 = -20 cm
The image will act as a real object for the convex lens
Applying les formula,
1/v1 - 1/u1 = 1/f1
Where, u1 = Object distance = -(20 +d) = -(20+8) = -20 -8 = -28,
v1 = Image distance
1/v2 =1/30 +1/-28 = 14-15/420 = -1/420
v2 = -420 cm
Hence, the parallel incident beam appears to diverge
that is (420 − 4)= 416 cm from the left of the centre of the combination of the two lenses.
(b) Height of the image, h1 = 1.5 cm
Object distance from convex lens =u1=-40cm
Lens formula :
1/f=1/v-1/u
1/v1=1/30 + 1/-40
=4-3/120
=1/120
V1=120cm
Magnification m = |v/u|=120/40=3
Hence m = 3 due to convex lens
The image formed by the convex lens acts as an object for the concave lens. According to the lens formula:
1/v2-1/u2=1/f2
Where
U2= object distance
=+(120-8) =112cm
V2= image distance
1/V2=1/-20 +1/112
= -112+20/2240
=-92/2240
V2= - 2240/92
Magnification = m1= |v2/u2|= 2240/92x 1/ 112
=20/92
Combined magnification = mx m1= 3 x 20/92
=60/92
=0.52
Combined magnification =0.652 = hi/h0
hi= image size
ho=object height
hi= 0.652x 1.5= 0.98cm
∴Size of image is 0.98cm
Answer:
(a)
(i) Let a parallel beam be the incident from the left on the convexlens first.
f 1 =30cm and u 1 =∞, give v 1 =+30cm.
This image becomes a virtual object for the second lens.
f 2 =20cm,u 2 =+(308)cm=+22cm which gives, v 2 =220cm.
The parallel incident beam appears to diverge from a point 216 cm from the centre of the two - lens system.
(ii) Let the parallel beam be incident from the left on the concavelens first:
f 1 =20cm,u 1 =∞, give v 1 =20cm.
This image becomes a real object for the second lens:
f 2 =+30cm,u 2 =(20+8)cm=28cm which gives, v 2 =420cm.
The paralle lincident beam appears to diverge from a point 416 cm on the left of the centre of the two - lens system.
Clearly, the answer depends on which side of the lens system the parallel beam is incident. Further we do not have a simple lens equation true for all u (and v) in terms of a definite constant of the system (the constant being determined by f 1 and f 2 , and the separation between the lenses). The notion of effective focal length, therefore, does not seem to be meaningful for this system.
(b)
u 1 =40cm,f 1 =30cm, gives v 1 =120cm.
Magnitude of magnification due to the first (convex) lens is 3.
u 2 =+(120)cm=+112cm (object virtual);
f 2
=20cm which gives v 2 = - (112×20 ) / 92
Magnitude of magnification due to the second (concave) lens = 20/92.
Net magnitude of magnification = 0.652
Size of the image = 0.98 cm