Question

An object is located $4\,m$ from the first of two thin converging lenses of focal lengths $2\,m$ and $1\,m$ respectively. The lenses are separated by $3\,m$. The final image formed by the second lens is located from the source at a distance of

• A. 8.0 m
• B. 5.5 m
• C. 6.0 m
• D. 6.5 m

Answer 1

Answer: (B)

From the lens formula (for first lens)
$\frac{1}{f_{1}} =\frac{1}{v_{1}}-\frac{1}{u_{1}}$
$\Rightarrow \frac{1}{2}=\frac{1}{v_{1}}-\frac{1}{(-4)}$
$\Rightarrow \frac{1}{2}+\frac{1}{4} =\frac{1}{v_{1}}=\frac{3}{4}$
$\Rightarrow v_{1} =\frac{4}{3}, u_{2}=3-\frac{4}{3}=5 / 3$
This image will be treated as the source for second lens, then again from lens formula, we have
$\frac{1}{f_{2}}=\frac{1}{v_{2}}-\frac{1}{u_{2}}$
$\Rightarrow \frac{1}{1}=\frac{1}{v_{2}}+5 / 3$ [By Eq. (i)]
$\Rightarrow 1-5 / 3=\frac{1}{v_{2}}$
$\Rightarrow v_{2}=-3 / 2$
This is the final image distance from 2 nd lens.
So, the overall distance of image from the primary source(or object)
Let $d=4+3-1.5=5.5$