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 40cm. Determine the magnification produced by the two-lens system, and the size of the image.
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
Height of the image, h1 = 1.5 cm
Object distance from the side of the convex lens, u1 = -40 cm
|u1| = 40cm
According to the lens formula:
…(1)
Where,
u1 = Object distance
v1 = Image distance
f1 = Focal length of convex lens.
By putting the values in equation (1), we get,
By solving the above equation we get,
v1 = 120 cm
Magnification, m = |v1|/|u1| = 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:
…(2)
Where,
u2 = Object distance
U2 = + (120 − 8) = 112 cm.
v2 = Image distance
f2 = Focal length of convex lens.
By putting the values in equation (2), we get,
By solving the above equation, we get,
v2 = -2240/92
magnification, m’ = |v2/u2| = 2240/92 × 1/112 = 20/92
Hence, the magnification due to the concave lens is 2092.
The magnification produced by the combination of the two lenses is calculated as:
M = m × m’ = 0.652
The magnification of the combination is given as:
h2/h1 = 0.652
h2 = h1 × 0.652
Where,
h1 = Object size = 1.5 cm
h2 = Size of the image
Therefore h2 = 0.652 × 1.5 = 0.98cm
Hence, the height of the image is 0.98 cm.