Question

A convex lens of focal length 10 cm, a concave lens of focal length 15 cm, and a third lens of unknown focal length are placed coaxially in contact. If the focal length of the combination is +12 cm, find the nature and focal length of the third lens, if all lenses are thin. Will the answer change if the lenses were thick?

Hint:

For thin lenses in contact, simply add the powers (\( \frac{1}{f} \)). For thick lenses, account for the separations between principal planes, which introduces additional terms in the combined focal length formula.

Answer 1

Part 1: Thin lenses Part 1: Thin Lenses
Step 1: List the given data.
- Convex lens: \( f_1 = 10 \, \text{cm} \) (positive, converging).
- Concave lens: \( f_2 = -15 \, \text{cm} \) (negative, diverging).
- Third lens: \( f_3 \), unknown.
- Combined focal length: \( F = 12 \, \text{cm} \) (positive, converging). For thin lenses in contact, the reciprocal of the combined focal length is the sum of the reciprocals of the individual focal lengths: \[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} \] Step 2: Substitute the values.
\[ \frac{1}{12} = \frac{1}{10} + \frac{1}{-15} + \frac{1}{f_3} \] Step 3: Solve for \( \frac{1}{f_3} \).
\[ \frac{1}{12} = 0.0833 \, \text{cm}^{-1} \] \[ \frac{1}{10} = 0.1 \, \text{cm}^{-1}, \quad \frac{1}{-15} = -0.0667 \, \text{cm}^{-1} \] \[ \frac{1}{10} + \frac{1}{-15} = 0.1 - 0.0667 = 0.0333 \, \text{cm}^{-1} \] \[ 0.0833 = 0.0333 + \frac{1}{f_3} \] \[ \frac{1}{f_3} = 0.0833 - 0.0333 = 0.05 \, \text{cm}^{-1} \] \[ f_3 = \frac{1}{0.05} = 20 \, \text{cm} \] Step 4: Determine the nature of the third lens.
Since \( f_3 = 20 \, \text{cm} \) is positive, the third lens is converging, i.e., a convex lens. Step 5: Verify.
\[ \frac{1}{F} = \frac{1}{10} + \frac{1}{-15} + \frac{1}{20} = 0.1 - 0.0667 + 0.05 = 0.0833 \] \[ F = \frac{1}{0.0833} \approx 12 \, \text{cm} \] This matches the given combined focal length, confirming the solution. % Part 2: Thick lenses Part 2: Effect of Thick Lenses
For thin lenses, we assume negligible thickness, so the lenses are effectively at the same position, and their powers add directly. For thick lenses in contact:
- The principal planes of each lens are separated by small distances \( d_1 \) (between the first and second lens) and \( d_2 \) (between the second and third lens), due to the thickness of the lenses.
- The formula for two lenses separated by distance \( d \) is: \[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \] For three lenses, the formula becomes complex, involving terms like \( -\frac{d_1}{f_1 f_2} \) and \( -\frac{d_2}{f_2 f_3} \). Without the thicknesses, we cannot compute the exact effect, but the focal length \( f_3 \) would need adjustment to account for these separations. The nature (convex) is likely to remain the same since the combination is converging, but the exact focal length will differ. % Final Answer Final Answer:
- For thin lenses, the third lens is convex with a focal length of \( 20 \, \text{cm} \).
- For thick lenses, the answer will change due to the effective separations between the principal planes, altering the focal length of the third lens, though the nature may remain converging.