Question

An object is located $4m$ from the first of two thin converging lenses of focal lengths $2m$ and $1m$ respectively. The lenses are separated by $3m$. 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 $d=4+3-1.5=5.5$