A thin convex lens of focal length 5 cm and a thin concave lens of focal length 4 cm are combined together (without any gap) and this combination has magnification m1 when an object is placed 10 cm before the convex lens. Keeping the positions of convex lens and object undisturbed a gap of 1 cm is introduced between the lenses by moving the concave lens away, which leads to a change in magnification of total lens system to m2. The value of | m1m2 | is .
- 2527
- 56
- 527
- 32
Answer
(2) 56
Solution
Lenses in contact: 1F = 15 - 14 = - 120
F = -20 cm
u = -10 cm: 1v = 1F + 1u = - 120 - 110 = - 320
v = - 203 cm
m1 = vu = -20/3-10 = 23
With a gap of 1 cm:
Convex lens: 1v1 = 15 + 1-10 = 110
v1 = 10 cm, m' = 10-10 = -1
This image lies 9 cm beyond the concave lens, i.e. a virtual object: u2 = +9 cm
Concave lens: 1v2 = - 14 + 19 = - 536
v2 = - 365 cm
m'' = v2u2 = -36/59 = - 45
m2 = m' m'' = (-1) (- 45 ) = 45
| m1m2 | = 23 54 = 56
Answer: (2)