Question

Figure shows an object $AB$ placed in front of two thin coaxial lenses $1$ and $2$ with focal lengths $24\, cm$ and $9.0 \,cm$ , respectively. The object is $6.0\, cm$ from the lens I and the lens separation is $L = 10\, cm$ . Where does the system of two lenses produce an image of the object $AB$ ?

• A. $+ 18 \, cm$
• B. - $18 \,cm$
• C. $+ 24 \, cm$
• D. $-24 \, cm$

Answer 1

Answer: (A)

$+ 18 \, cm$

Answer 2

For first lens $f_{1}=24.0\,cm$
$u_{1}=6.0\,cm$
To find $v_{1}$
Using lens formula $\frac{1}{v}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$
$\frac{1}{v}=\frac{1}{f_{1}}+\frac{1}{u_{1}}=\frac{1}{24}-\frac{1}{6}$
$\frac{1}{v}=\frac{1-4}{24}=-\frac{3}{24}$
$v=-8\,cm$
$\therefore$ For second lens
$u_{2}=-8\,cm+\left(-10\,cm\right)=-18\,cm$
$f_{2}=9\,cm$
Again using lens formula
$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$
$\frac{1}{v}=\frac{1}{f_{2}}+\frac{1}{u_{2}}=\frac{1}{9}-\frac{1}{18}$
$=\frac{2-1}{18}=\frac{1}{18}$
$\Rightarrow v=18\,cm$
$\therefore$ Final image will be formed at $18\, cm$ to the right of second lens