A thin convex lens of focal length 5 cm and a thin concave lens of focal length 4 cm are combined to

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 $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

1. Initial State (Lenses in contact to find $m_1$ )
- Equivalent focal length $(F): \frac{1}{F}=\frac{1}{5}-\frac{1}{4} \Longrightarrow F=-20 \mathrm{~cm}$
- Using magnification formula $m=\frac{F}{F+u}$ where $u=-10 \mathrm{~cm}$ 
$
m_1=\frac{-20}{-20+(-10)}=\frac{2}{3}
$
2. Final State ( 1 cm gap introduced to find $m_2$ )
- Convex lens: $m=\frac{f}{f+u} \Longrightarrow m_{	ext {convex }}=\frac{5}{5+(-10)}=-1$.
- Since $m=\frac{v}{u}$, the image forms at $v=+10 \mathrm{~cm}$.
- Concave lens: The gap is 1 cm , so the first image acts as a virtual object at $u=10- 1=+9 \mathrm{~cm}$.
- $m_{	ext {concave }}=\frac{f}{f+u}=\frac{-4}{-4+9}=-\frac{4}{5}$.
- Total new magnification ( $m_2$ ):
$
m_2=m_{	ext {convex }} 	imes m_{	ext {concave }}=(-1) 	imes\left(-\frac{4}{5}ight)=\frac{4}{5}
$
3. Required Ratio
$
\left|\frac{m_1}{m_2}ight|=\left|\frac{2 / 3}{4 / 5}ight|=\frac{2}{3} 	imes \frac{5}{4}=\frac{\mathbf{5}}{\mathbf{6}}
$