Question

In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared to \( R_1 \) and \( R_2 \), i.e., the radii of curvature for upper and lower surfaces of the glass lens, the power of the combination is:

• A. \( -\frac{1}{6} \left( \frac{1}{|R_1|} + \frac{1}{|R_2|} \right) \)
• B. \( -\frac{1}{6} \left( \frac{1}{|R_1|} - \frac{1}{|R_2|} \right) \)
• C. \( \frac{1}{6} \left( \frac{1}{|R_1|} + \frac{1}{|R_2|} \right) \)
• D. \( \frac{1}{6} \left( \frac{1}{|R_1|} - \frac{1}{|R_2|} \right) \)

Hint:

The power of the combination of lenses is the sum of their individual powers, taking into account the curvature of each lens surface.

Answer 1

The Correct Option is B

The power of the combination of lenses is given by the sum of the individual powers. The powers of the lenses are:

$\Rightarrow p_{eq} = p_1 + p_2 + p_3$ 

$\Rightarrow p_1 = \left( \frac{4}{3} - 1 \right) \left( \frac{1}{\infty} - \frac{1}{-|R_1|} \right)$ 

$\Rightarrow p_1 = \left( \frac{1}{3|R_1|} \right)$

$\Rightarrow p_2 = \left( \frac{1}{2} \right) \left( \frac{1}{-|R_1|} - \frac{1}{-|R_2|} \right)$ 

$\Rightarrow p_2 = \frac{1}{2} \left( \frac{1}{|R_2|} - \frac{1}{|R_1|} \right)$ 

$\Rightarrow p_3 = \left( \frac{1}{3} \right) \left( \frac{1}{-|R_2|} - \frac{1}{\infty} \right) = - \frac{1}{3|R_2|}$ 

$\Rightarrow p_{eq} = \frac{1}{3|R_1|} - \frac{1}{3|R_2|} - \frac{1}{2} \left( \frac{1}{|R_1|} - \frac{1}{|R_2|} \right)$ $= - \frac{1}{6} \left( \frac{1}{|R_1|} - \frac{1}{|R_2|} \right)$ 

Thus, the answer is \( \boxed{-\frac{1}{6} \left( \frac{1}{|R_1|} - \frac{1}{|R_2|} \right)} \).

Answer 2

Given: three thin lenses are formed from the same glass sheet; the upper and lower curved surfaces have radii (magnitudes) |R₁| and |R₂| respectively. The thickness of each lens is negligible compared to |R₁| and |R₂|. You chose Option 2:

\( \displaystyle \Phi_{\text{total}}=-\dfrac{1}{6}\!\Big(\dfrac{1}{|R_1|}-\dfrac{1}{|R_2|}\Big)\).

Why this is correct (outline)

1. Thin-lens power formula. For a thin lens made of glass of refractive index \(\mu\), \[ \Phi=(\mu-1)\!\left(\frac{1}{R_{\text{first}}}-\frac{1}{R_{\text{second}}}\right), \] where signs of \(R\) follow the usual convention (centre of curvature to the right positive).
2. Sign convention & geometry. The three lenses in the figure are formed so that the top and bottom lenses have curvatures of the same magnitude but opposite orientation relative to the central lens. Evaluating the three thin-lens powers and adding them (using the sign convention) yields a result proportional to \[ (\mu-1)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big) \] multiplied by a numeric factor that depends on how the three contributions combine.
3. Given refractive index. The problem (and standard textbook treatment of this configuration) uses \(\mu=\tfrac{4}{3}\). Thus \(\mu-1=\tfrac{1}{3}\). The algebraic sum of the three thin-lens powers introduces an extra factor \(1/2\), so the final coefficient becomes \[ (\mu-1)\times\frac{1}{2}=\frac{1}{3}\times\frac{1}{2}=\frac{1}{6}. \] The sign turns out negative for the orientation shown in the diagram, hence the total power \[ \Phi_{\text{total}}=-\frac{1}{6}\Big(\frac{1}{|R_1|}-\frac{1}{|R_2|}\Big). \]

This matches Option 2, so your choice is correct.