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

The Correct Option is D

$f _{1}=+ f _{1}$
$f _{2}=+ f _{2}$
$f _{3}=- f _{3}$
$\frac{1}{F_{R}}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$
$P_{R}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$
$P_{1}=\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{f_{2}+f_{1}}{f_{1} f_{2}}$
$P_{2}=\frac{1}{f_{1}}+\frac{1}{f_{3}}=\frac{f_{3}-f_{1}}{f_{1} f_{3}}$
$P_{2}=\frac{1}{f_{2}}+\frac{1}{f_{3}}=\frac{f_{3}-f_{2}}{f_{2} f_{3}}$