Question
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 $m_{1}$ 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 lead to a change in magnification of total lens system to $m_{2}$. The value of $\left| \frac{m_{1}}{m_{2}} \right|$ is $\_\_\_\_$ .
Options
Solution
In contact:
1/F = 1/5 − 1/4 = −1/20 → F = −20 cm
u = −10 cm → 1/v = 1/F + 1/u = −1/20 − 1/10 = −3/20 → v = −20/3 cm
m₁ = v/u = (20/3)/10 = 2/3
With 1 cm gap:
Convex: u = −10, f = 5 → v₁ = 10 cm (real image 10 cm right of convex)
For concave (1 cm after convex): object distance u₂ = +(10 − 1) = +9 cm
1/v₂ = 1/(−4) + 1/9 = −5/36 → v₂ = −36/5 cm
m_convex = 10/(−10) = −1 m_concave = (−36/5)/9 = −4/5
m₂ = (−1) × (−4/5) = 4/5
|m₁ / m₂| = (2/3) ÷ (4/5) = (2/3) × (5/4) = 10/12 = 5/6
Create a free account to view solution
View Solution Free