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Difficulty: HardThin Lenses, Optical Instruments, and Defects of Vision

A converging lens of focal length 20 cm20\text{ cm} is placed in thin coaxial contact with a diverging lens of focal length 50 cm50\text{ cm}. An object is placed 30 cm30\text{ cm} in front of this lens combination. What is the position and nature of the final image formed?

  1. 300 cm300\text{ cm} in front of the combination (virtual image)Answer
  2. B
    60 cm60\text{ cm} behind the combination (real image)
  3. C
    27.3 cm27.3\text{ cm} behind the combination (real image)
  4. D
    15.8 cm15.8\text{ cm} behind the combination (real image)

Answer

The image is virtual and formed 300 cm300\text{ cm} in front of the lens combination.
Combining a converging lens (f=+20 cmf = +20\text{ cm}) and a diverging lens (f=−50 cmf = -50\text{ cm}) yields an effective focal length of F=+1003 cmF = +\frac{100}{3}\text{ cm}. Using the lens formula 1F=1u+1v\frac{1}{F} = \frac{1}{u} + \frac{1}{v} with object distance u=30 cmu = 30\text{ cm} gives 1v=3100−130=−1300 cm−1\frac{1}{v} = \frac{3}{100} - \frac{1}{30} = -\frac{1}{300}\text{ cm}^{-1}, resulting in v=−300 cmv = -300\text{ cm}. The negative sign confirms the image is virtual and located 300 cm300\text{ cm} in front of the combination.

Step-by-Step Solution

1
Calculate the effective focal length (FF) of the two lenses in contact.
1F=1f1+1f2=120 cm+1−50 cm=5−2100 cm=3100 cm−1\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{20\text{ cm}} + \frac{1}{-50\text{ cm}} = \frac{5 - 2}{100\text{ cm}} = \frac{3}{100}\text{ cm}^{-1}, so F=+1003 cmF = +\frac{100}{3}\text{ cm}.
Thin lenses in contact combine algebraically according to their optical powers, taking signs into account (positive for converging, negative for diverging).
2
Apply the lens formula to find the image distance (vv).
1F=1u+1v  ⟹  3100=130+1v  ⟹  1v=3100−130=9−10300=−1300 cm−1\frac{1}{F} = \frac{1}{u} + \frac{1}{v} \implies \frac{3}{100} = \frac{1}{30} + \frac{1}{v} \implies \frac{1}{v} = \frac{3}{100} - \frac{1}{30} = \frac{9 - 10}{300} = -\frac{1}{300}\text{ cm}^{-1}.
Rearranging the thin lens equation allows us to solve for the image distance vv given object distance u=30 cmu = 30\text{ cm}.
3
Interpret the sign and magnitude of vv.
v=−300 cmv = -300\text{ cm}, which signifies a virtual image located 300 cm300\text{ cm} in front of the lens combination (on the object side).
A negative image distance in the standard real-is-positive convention denotes a virtual image.

Key Concept

Combination of thin lenses in contact and lens sign conventions
Estimated Time:2m 0s
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Topics

Thin Lenses, Optical Instruments, and Defects of Vision

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