Question

Two thin biconvex lenses have focal lengths f₁ and f₂. A third thin biconcave lens has focal length of f₃. If the two biconvex lenses are in contact, the total power of the lenses is P₁. If the first convex lens is in contact with the third lens, the total power is P₂. If the second lens is in contact with the third lens, the total power is P₃ then

• A. \(P_1=\frac{f_1f_2}{f_1-f_2},P_2=\frac{f_1f_3}{f_3-f_1}\ \text{and}\ P_3=\frac{f_2f_3}{f_3-f_2}\)
• B. \(P_1=\frac{f_1-f_2}{f_1f_2},P_2=\frac{f_3-f_1}{f_3+f_1}\ \text{and}\ P_3=\frac{f_3-f_2}{f_2f_3}\)
• C. \(P_1=\frac{f_1-f_2}{f_1f_2},P_2=\frac{f_3-f_1}{f_1f_3}\ \text{and}\ P_3=\frac{f_3-f_2}{f_2f_3}\)
• D. \(P_1=\frac{f_1+f_2}{f_1f_2},P_2=\frac{f_3-f_1}{f_1f_3}\ \text{and}\ P_3=\frac{f_3-f_2}{f_2f_3}\)

Answer 1

The Correct Option is D

The power ($P$) of a lens is the reciprocal of its focal length ($f$): $P = \frac{1}{f}$. When two thin lenses are in contact, the total power is the sum of their individual powers: $P_{total} = P_1 + P_2$. A biconcave lens has a negative focal length.

Let:

• $P_1 = \frac{1}{f_1}$
• $P_2 = \frac{1}{f_2}$
• $P_3 = \frac{1}{f_3}$

Case 1: Two biconvex lenses in contact:

$P = P_1 + P_2$

$P_1 = \frac{1}{f_1} + \frac{1}{f_2} = \frac{f_1 + f_2}{f_1 f_2}$

Case 2: First convex lens and biconcave lens in contact:

$P = P_1 + P_3$

$P_2 = \frac{1}{f_1} + \frac{1}{f_3} = \frac{f_3 + f_1}{f_1 f_3}$

Case 3: Second convex lens and biconcave lens in contact:

$P = P_2 + P_3$

$P_3 = \frac{1}{f_2} + \frac{1}{f_3} = \frac{f_3 + f_2}{f_2 f_3}$

We can express these as:

• $P_1 = \frac{f_1+f_2}{f_1 f_2}$
• $P_2 = \frac{f_3+f_1}{f_1 f_3}$
• $P_3 = \frac{f_3+f_2}{f_2 f_3}$

$P_1= \frac{f_1+f_2}{f_1 f_2}$, $P_2=\frac{f_3+f_1}{f_1 f_3}$ and $P_3= \frac{f_3+f_2}{f_2 f_3}$ The correct answer is (D) .

Answer 2

For thin lenses in contact, the power of the combination is the algebraic sum of the powers of the individual lenses. The power of a lens $P$ is the reciprocal of its focal length $f$ ($P = 1/f$).

For biconvex lenses, focal lengths $f_1$ and $f_2$ are positive. For a biconcave lens, focal length $f_3$ is negative. Let's consider $f_3$ as the magnitude and use $-f_3$ for power calculation.

Case 1: Two biconvex lenses are in contact.

The total power $P_1$ is the sum of the powers of the two biconvex lenses:

$P_1 = P_{\text{lens1}} + P_{\text{lens2}} = \frac{1}{f_1} + \frac{1}{f_2}$

$P_1 = \frac{f_1 + f_2}{f_1 f_2}$

Case 2: The first convex lens is in contact with the third (biconcave) lens.

The total power $P_2$ is the sum of the powers of the first convex lens and the biconcave lens. Since the focal length of the biconcave lens is effectively $-f_3$ (considering $f_3$ as magnitude), the power of the biconcave lens is $-\frac{1}{f_3}$.

$P_2 = P_{\text{lens1}} + P_{\text{lens3}} = \frac{1}{f_1} - \frac{1}{f_3}$

$P_2 = \frac{f_3 - f_1}{f_1 f_3}$

Case 3: The second convex lens is in contact with the third (biconcave) lens.

The total power $P_3$ is the sum of the powers of the second convex lens and the biconcave lens.

$P_3 = P_{\text{lens2}} + P_{\text{lens3}} = \frac{1}{f_2} - \frac{1}{f_3}$

$P_3 = \frac{f_3 - f_2}{f_2 f_3}$

Comparing Derived Expressions with Options:

Option (A) \(P_1=\frac{f_1f_2}{f_1-f_2},P_2=\frac{f_1f_3}{f_3-f_1}\ \text{and}\ P_3=\frac{f_2f_3}{f_3-f_2}\) is incorrect for $P_1$.

Option (B) \(P_1=\frac{f_1-f_2}{f_1f_2},P_2=\frac{f_3-f_1}{f_3+f_1}\ \text{and}\ P_3=\frac{f_3-f_2}{f_2f_3}\) is incorrect for $P_1$ and $P_2$.

Option (C) \(P_1=\frac{f_1-f_2}{f_1f_2},P_2=\frac{f_3-f_1}{f_1f_3}\ \text{and}\ P_3=\frac{f_3-f_2}{f_2f_3}\) is incorrect for $P_1$.

Option (D) \(P_1=\frac{f_1+f_2}{f_1f_2},P_2=\frac{f_3-f_1}{f_1f_3}\ \text{and}\ P_3=\frac{f_3-f_2}{f_2f_3}\) is correct for all $P_1$, $P_2$, and $P_3$.

Final Answer: The final answer is ${P_1=\frac{f_1+f_2}{f_1f_2},P_2=\frac{f_3-f_1}{f_1f_3}\ \text{and}\ P_3=\frac{f_3-f_2}{f_2f_3}}$