Related Formula
Lens Maker's Formula:
(1)/(f) = (μ - 1) ((1)/(R₁) - (1)/(R₂))$$\frac{1}{f} = (\mu - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
For a bi-convex lens of equal radii (R₁ = +R$R_1 = +R$, R₂ = -R$R_2 = -R$):
(1)/(f) = (μ - 1) (2)/(R)$$\frac{1}{f} = (\mu - 1) \frac{2}{R}$$
Power P ∝ (1)/(f)$P \propto \frac{1}{f}$.
Core Logic
For the initial lens:
- R = (1)/(6) cm$R = \frac{1}{6} \ \mathrm{cm}$
- (1)/(f₁) = (μ - 1) (2)/(1/6) = 12 (μ - 1)$\frac{1}{f_1} = (\mu - 1) \frac{2}{1/6} = 12 (\mu - 1)$
For the replacement lens (R₁ = +R₁$R_1 = +R_1$, R₂ = -R₂$R_2 = -R_2$):
(1)/(f₂) = (μ - 1) ((1)/(R₁) + (1)/(R₂))$$\frac{1}{f_2} = (\mu - 1) \left(\frac{1}{R_1} + \frac{1}{R_2}\right)$$
Since power must be preserved (f₁ = f₂$f_1 = f_2$):
(1)/(R₁) + (1)/(R₂) = (2)/(R) = 12 cm⁻¹$$\frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{R} = 12 \ \mathrm{cm}^{-1}$$
We need to find a combination where the sum of the reciprocals of the radii equals 12$12$.
Step 1: Verify the options
Let's check each choice:
- Option (1): R₁ = (1)/(3)$R_1 = \frac{1}{3}$, R₂ = (1)/(3)$R_2 = \frac{1}{3}$:
(1)/(1/3) + (1)/(1/3) = 3 + 3 = 6 ≠ 12$$\frac{1}{1/3} + \frac{1}{1/3} = 3 + 3 = 6 \neq 12$$
- Option (2): R₁ = (1)/(5)$R_1 = \frac{1}{5}$, R₂ = (1)/(7)$R_2 = \frac{1}{7}$:
(1)/(1/5) + (1)/(1/7) = 5 + 7 = 12 (Correct Combination)$$\frac{1}{1/5} + \frac{1}{1/7} = 5 + 7 = 12 \quad \checkmark \quad \text{(Correct Combination)}$$
- Option (3): R₁ = (1)/(3)$R_1 = \frac{1}{3}$, R₂ = (1)/(7)$R_2 = \frac{1}{7}$:
(1)/(1/3) + (1)/(1/7) = 3 + 7 = 10 ≠ 12$$\frac{1}{1/3} + \frac{1}{1/7} = 3 + 7 = 10 \neq 12$$
- Option (4): R₁ = (1)/(6)$R_1 = \frac{1}{6}$, R₂ = (1)/(9)$R_2 = \frac{1}{9}$:
(1)/(1/6) + (1)/(1/9) = 6 + 9 = 15 ≠ 12$$\frac{1}{1/6} + \frac{1}{1/9} = 6 + 9 = 15 \neq 12$$
Thus, only Option (2) meets the physical conditions.
Pattern Recognition
Sees: Equivalent thin lens power with modified surfaces.
Trap: Neglecting sign convention for the second spherical surface during substitution.
Shortcut: If the radii are of the form 1/n$1/n$, then the sum of n₁ + n₂$n_1 + n_2$ must equal 2 × ninitial$2 \times n_{\text{initial}}$. Since initial n = 6$n = 6$, 2 × 6 = 12$2 \times 6 = 12$. The only pairing whose denominators add up to 12 is 5 + 7$5 + 7$.
Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments