Question

In the displacement method, a concave lens is placed in between an object and a screen. If the magnification in the two positions are m${}_1$ and m${}_2$ (m${}_1$ > m${}_2$), and the distance between the two positions of the lens is x, the focal length of the lens is

• A. $\frac{x}{m_1+m_2}$
• B. $\frac{x}{m_1-m_2}$
• C. $\frac{x}{(m_1+m_2{)}^2}$
• D. $\frac{x}{(m_1-m_2{)}^2}$

Answer 1

The correct option is A $\frac{x}{m_1+m_2}$

For a lens,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$⟹f=\frac{uv}{u-v}$

In the displacement method, $x=v+u$

$m_1=\frac{v}{u}$,$m_2=\frac{u}{v}$

$m_2-m_1=\frac{u^2-v^2}{uv}$

$⟹f(m_2-m_1)=u+v=x$

Hence, $f=\frac{x}{m_2-m_1}$

But for a concave lens, the focus is on the left side of the lens, hence negative by convention.

Thus the focal length of the concave lens is $\frac{x}{m_1-m_2}$