Question

Two thin biconvex lenses have focal lengths $f_1$ and $f_2.$ A third thin biconcave lens has 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. $P_1=\frac{f_1{f}_2}{f_1-f_2},P_2=\frac{f_1{f}_3}{f_3-f_1}$ and $P_3=\frac{f_2{f}_3}{f_3-f_2}$
• B. $P_1=\frac{f_1-f_2}{f_1{f}_2},P_2=\frac{f_3-f_1}{f_3+f_1}$ and $P_3=\frac{f_3-f_2}{f_2{f}_3}$
• C. $P_1=\frac{f_1-f_2}{f_1{f}_2},P_2=\frac{f_3-f_1}{f_1{f}_3}$ and $P_3=\frac{f_3-f_2}{f_2{f}_3}$
• D. $P_1=\frac{f_1+f_2}{f_1{f}_2},P_2=\frac{f_3-f_1}{f_1{f}_3}$ and $P_3=\frac{f_3-f_2}{f_2{f}_3}$

Answer 1

Answer: (D)

$f_1=+f_1$
$f_2=+f_2$
$f_3=-f_3$
$\frac{1}{F_R}=\frac{1}{f_1}+\frac{1}{f_2}$
$P_R=\frac{1}{f_1}+\frac{1}{f_2}$
$P_1=\frac{1}{f_1}+\frac{1}{f_2}=\frac{f_2+f_1}{f_1{f}_2}$
$P_2=\frac{1}{f_1}+\frac{1}{f_3}=\frac{f_3-f_1}{f_1{f}_3}$
$P_2=\frac{1}{f_2}+\frac{1}{f_3}=\frac{f_3-f_2}{f_2{f}_3}$