Question

An object is located 4m from the first of two thin converging lenses of local lengths 2m and 1m 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 source.

• A. $8.0m$
• B. $7.5m$
• C. $6.0m$
• D. $6.5m$

Answer 1

The correct option is B $7.5m$

Let the position of image formed after refraction from first lens be $v_1$.

Hence from lens equation

$\frac{1}{v_1}-\frac{1}{-4}=\frac{1}{2}$

Hence $v_1=4m$

Thus this image, which acts as object for second refraction, is $1m$ ahead of the second lens.

Hence $u_2=+1$

From lens equation for second lens,

$\frac{1}{v}-\frac{1}{u_2}=\frac{1}{f_2}$

$⟹\frac{1}{v}-\frac{1}{1}=\frac{1}{1}$

$⟹v=0.5m$

Hence distance from image from the source=$4m+3m+0.5m=7.5m$