Question

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

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

Answer 1

Answer: (A)

For first lens $f_1=24.0cm$
$u_1=6.0cm$
To find $v_1$
Using lens formula $\frac{1}{v}-\frac{1}{u_1}=\frac{1}{f_1}$
$\frac{1}{v}=\frac{1}{f_1}+\frac{1}{u_1}=\frac{1}{24}-\frac{1}{6}$
$\frac{1}{v}=\frac{1-4}{24}=-\frac{3}{24}$
$v=-8cm$
$\therefore$ For second lens
$u_2=-8cm+\left(-10cm\right)=-18cm$
$f_2=9cm$
Again using lens formula
$\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2}$
$\frac{1}{v}=\frac{1}{f_2}+\frac{1}{u_2}=\frac{1}{9}-\frac{1}{18}$
$=\frac{2-1}{18}=\frac{1}{18}$
$\Rightarrow v=18cm$
$\therefore$ Final image will be formed at $18cm$ to the right of second lens