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

To determine the total power of lenses in contact, we use the fundamental formula for lens power:

\(P = \frac{1}{f}\) 

Where \(P\) is the power and \(f\) is the focal length of the lens.

When multiple lenses are in contact, their combined power is the sum of their individual powers:

\(P_{total} = P_1 + P_2 + P_3 + \ldots\)

1. Analyzing the given situation:
   • For the combination of two biconvex lenses with focal lengths \(f_1\) and \(f_2\), the total power is given by:

\(P_1 = \frac{1}{f_1} + \frac{1}{f_2} = \frac{f_1 + f_2}{f_1 f_2}\)

• For the combination of the first biconvex and the biconcave lens with focal lengths \(f_1\) and \(f_3\) respectively, the total power is:

\(P_2 = \frac{1}{f_1} + \frac{1}{-f_3} = \frac{f_3 - f_1}{f_1 f_3}\)

• For the combination of the second biconvex and the biconcave lens with focal lengths \(f_2\) and \(f_3\) respectively, the total power is:

\(P_3 = \frac{1}{f_2} + \frac{1}{-f_3} = \frac{f_3 - f_2}{f_2 f_3}\)

1. Justifying the Correct Option:
   • The derived expressions for the total powers \(P_1\), \(P_2\), and \(P_3\) match the given correct answer option:
   • \(P_1 = \frac{f_1 + f_2}{f_1 f_2}\)
   • \(P_2 = \frac{f_3 - f_1}{f_1 f_3}\)
   • \(P_3 = \frac{f_3 - f_2}{f_2 f_3}\)

Conclusion: Thus, based on the above analysis and calculations, the correct answer is \(P_{1}=\frac{f_{1}+f_{2}}{f_{1} f_{2}}, P_{2}=\frac{f_{3}-f_{1}}{f_{1} f_{3}}\) and \(P_{3}=\frac{f_{3}-f_{2}}{f_{2} f_{3}}\).

Answer 2

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