Question

A lens of focal length ${}^{\prime }{f}^{\prime }$ is placed in between an object and screen at a distance ${}^{\prime }{D}^{\prime }$. The lens forms two real images of object on the screen for two of its different positions, a distance ${}^{\prime }{x}^{\prime }$ apart. The two real images have magnifications $m_1$ and $m_2$, respectively $(m_1>m_2)$. Then

• A. $f=x/(m_1-m_2)$
• B. $m_1{m}_2=1$
• C. $f=(D^2-x^2)/4D$
• D. $D\ge 4f$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $m_1{m}_2=1$
C $f=(D^2-x^2)/4D$
D $D\ge 4f$

$v+u=D$
and $v-u=x\Rightarrow v=\frac{D+x}{2},u=\frac{D-u}{2}$ and $f=\frac{D^2-x^2}{4D}$
$m_1=\frac{D+x}{D-x},m_2=\frac{D-x}{D+x}$.