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)

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