Question

(a)Determine the ‘effective focal length’ of the combination of the two lenses if they are placed 8.0cm 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 1.5cm 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 40cm. Determine the magnification produced by the two-lens system and the size of the image.

Answer 1

Focal length of the convex lens, f1 = 30cm
Focal length of the concave lens, f2 = -20cm
Distance between the two lenses, d = 8.0cm
(a) When the parallel beam of light is incident on the convex lens first:
According to the lens formula, we have:
$\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1}$
Where,
u1 = object distance = v1
= Image distance
$\frac{1}{v_1}=\frac{1}{30}-\frac{1}{\infty}=\frac{1}{30}$
∴ v1 = 30cm
The image will act as a virtual object for the concave lens.
Applying the lens formula to the concave lens, we have:
$\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2}$
Where,
u2 = Object distance
= (30-d) = 30-8 = 22cm
v2 = Image distance
$\frac{1}{v_2}=\frac{1}{22}-\frac{1}{20}=\frac{10-11}{220}=-\frac{1}{220}$
∴v2 = -220cm
The parallel incident beam appears to diverge from a point that is ($220-\frac{d}{2}$ = 220-4) 216cm from the center of the combination of the two lenses.
(ii) When the parallel beam of light is incident, from the left, on the concave lens first:
According to the lens formula, we have:
$\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2}$
Where,
u2 = object distance = -∞
v2=Image distance
$\frac{1}{v_2}=\frac{1}{-20}+\frac{1}{-\infty}=-\frac{1}{20}$
∴v2 = -20cm
The image will act as a real object for the convex lens.
Applying the lens formula to the convex lens, we have:
$\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1}$
Where,
u1 = object distance
= -(20+d) = -(20 +8) = -28cm
v1 = Image distance
$\frac{1}{v_1}=\frac{1}{30}+\frac{1}{-28}=\frac{14-15}{420}=-\frac{1}{420}$
∴ v2 = -420cm
Hence, the parallel incident beam appears to diverge from a point that is (420-4) 416cm from the left of the center of the combination of the two lenses. The answer does depend on the side of the combination at which the parallel beam of light is incident. The notion of effective focal length does not seem to be useful for this combination.
(b) Height of the image, h1 = 1.5cm
Object distance from the side of a convex lens, u1 = 40cm
|u1| = 40cm
According to the lens formula: $\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1}$
Where,
v1 = Image distance
$\frac{1}{v_1}=\frac{1}{30}+\frac{1}{-40}$ = $\frac{4-3}{120}$ = $\frac{1}{120}$
∴v1 = 120cm
Magnification, m=$\frac{v_1}{|u_1|}$= $\frac{120}{40}$ = 3
Hence, the magnification due to the convex lens is 3.
The image formed by the convex lens acts as an object for the concave lens.
According to the lens formula: $\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2}$
Where,
u2 = Object distance
= (120-8) = 112cm
v2 = Image distance
$\frac{1}{v_2}=\frac{1}{-20}+\frac{1}{112}$ = $\frac{-112+20}{2240}$= $-\frac{92}{2240}$
∴v2 =$-\frac{2240}{92}$ cm
Magnification, m' =|$\frac{v_2}{u_2}$|= $\frac{2240}{92}\times \frac{1}{112}$ = $\frac{20}{92}$
Hence, the magnification due to the concave lens is $\frac{20}{92}$.
The magnification produced by the combination of the two lenses is calculated as: m×m'
= $\frac{3\times20}{92}$ = $\frac{60}{92}$ = 0.652
The magnification of the combination is given as:
$\frac{h_2}{h_1}$ = 0.652
h2 = 0.652×h1
Where,
h1 = Object size = 1.5cm
h2 = Size of the image
∴ h2 = 0.652×1.5 = 0.98cm
Hence, the height of the image is 0.98cm.