Question

Curved surfaces of a plano-convex lens of refractive index \(\mu_1\) and a plano-concave lens of refractive index \(\mu_2\) have equal radius of curvature as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses. 

 

• A. \(\mu_1 - \mu_2\)
• B. \(\frac{1}{\mu_1 - \mu_2}\)
• C. \(\mu_2 - \mu_1\)
• D. \(\frac{1}{\mu_2 - \mu_1}\)

Hint:

Always be careful with the sign convention in the Lens Maker's formula. For a plano-convex lens, one radius is R and the other is \(\infty\). For a plano-concave lens, one is \(\infty\) and the other is -R (if the curved surface is the first one) or R (if it's the second). The final result is the same. Combining the powers (\(P=1/f\)) is often the quickest way for lens combinations.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We have a combination of a plano-convex and a plano-concave lens. We need to find the ratio of their common radius of curvature (R) to the equivalent focal length (F) of the combination.
Step 2: Key Formula or Approach:
1. Lens Maker's Formula: \( \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \)
2. Combination of Lenses: For two lenses in contact, the equivalent focal length F is given by \( \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \).
Step 3: Detailed Explanation:
Let R be the radius of curvature for both curved surfaces.
For the plano-convex lens (Lens 1):
- Refractive index = \(\mu_1\).
- For the curved surface, \(R_1 = R\). For the plane surface, \(R_2 = \infty\).
Using the Lens Maker's formula:
\[ \frac{1}{f_1} = (\mu_1 - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{\mu_1 - 1}{R} \] For the plano-concave lens (Lens 2):
- Refractive index = \(\mu_2\).
- For the plane surface, \(R_1 = \infty\). For the curved surface, \(R_2 = R\). A concave surface seen from the left has a positive radius of curvature according to the sign convention.
Using the Lens Maker's formula:
\[ \frac{1}{f_2} = (\mu_2 - 1) \left( \frac{1}{\infty} - \frac{1}{R} \right) = - \frac{\mu_2 - 1}{R} \] For the combination:
The equivalent focal length F is given by:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \] \[ \frac{1}{F} = \frac{\mu_1 - 1}{R} - \frac{\mu_2 - 1}{R} \] \[ \frac{1}{F} = \frac{(\mu_1 - 1) - (\mu_2 - 1)}{R} = \frac{\mu_1 - 1 - \mu_2 + 1}{R} = \frac{\mu_1 - \mu_2}{R} \] We need to find the ratio \( \frac{R}{F} \).
From the above equation, we can rearrange to find this ratio:
\[ \frac{R}{F} = \mu_1 - \mu_2 \] Step 4: Final Answer:
The ratio of the radius of curvature to the focal length of the combined lenses is \( \mu_1 - \mu_2 \).