In the given optical instrument,how can a third lens of the same focal length be inserted to increase the light-collecting efficiency without changing the positions of the object and the image? Calculate the position of that lens.

Two thin biconvex lenses have focal lengths $f_{1}$ and $f_{2}$. $A$ third thin biconcave lens has a focal length of $f_{3}$. If the two biconvex lenses are in contact,the total power of the lenses is $P_{1}$. If the first convex lens is in contact with the third lens,the total power is $P_{2}$. If the second lens is in contact with the third lens,the total power is $P_{3}$,then:

$A$ plano-convex lens of refractive index $\mu_1$ fits exactly into a plano-concave lens of refractive index $\mu_2$. Their plane surfaces are parallel to each other. $R$ is the radius of curvature of the curved surface of the lenses. The focal length of the combination is:

The position of the final image formed by the given lens combination from the third lens will be at a distance of $(f_1 = +10 \ cm, f_2 = -10 \ cm, f_3 = +30 \ cm)$.

$A$ bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. The refractive index $n$ of the first lens is $1.5$ and that of the second lens is $1.2$. Both the curved surfaces are of the same radius of curvature $R = 14 \, cm$. For this bi-convex lens,for an object distance of $40 \, cm$,the image distance will be.......$cm$

An object is placed $0.40 \,m$ from one of the two lenses $L_1$ and $L_2$ of focal lengths $0.20 \,m$ and $0.10 \,m$ respectively,as depicted in the figure. The separation between the lenses is $0.30 \,m$. The final image formed by this two-lens system is at