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Difficulty: Very hardThin Lenses, Optical Instruments, and Defects of Vision

A myopic person has a far point of 52 cm52\text{ cm} from the eye. A corrective lens is placed 2 cm2\text{ cm} in front of the eye to enable the person to see distant objects clearly. A second thin converging lens of focal length +20 cm+20\text{ cm} is then placed in thin coaxial contact with this corrective lens. If an object is placed 50 cm50\text{ cm} in front of this combined lens system, what is the position and nature of the final image formed?

  1. 100 cm100\text{ cm} behind the combined lens (real)Answer
  2. B
    20 cm20\text{ cm} behind the combined lens (real)
  3. C
    100 cm100\text{ cm} in front of the combined lens (virtual)
  4. D
    20 cm20\text{ cm} in front of the combined lens (virtual)

Answer

100 cm100\text{ cm} behind the combined lens (real image)
The corrective lens for short-sightedness (myopia) must be a diverging lens with a negative focal length. Accounting for the 2 cm2\text{ cm} distance between the eye and the lens, the far point relative to the lens is 50 cm50\text{ cm}, giving f1=−50 cmf_1 = -50\text{ cm}. Combining this lens with the converging lens (f2=+20 cmf_2 = +20\text{ cm}) yields a net power 1F=−150+120=+3100 cm−1\frac{1}{F} = -\frac{1}{50} + \frac{1}{20} = +\frac{3}{100}\text{ cm}^{-1}, or F=+1003 cmF = +\frac{100}{3}\text{ cm}. Placing an object at u=50 cmu = 50\text{ cm} gives 1v=3100−150=+1100 cm−1\frac{1}{v} = \frac{3}{100} - \frac{1}{50} = +\frac{1}{100}\text{ cm}^{-1}, which yields v=+100 cmv = +100\text{ cm}. The positive sign confirms a real image formed 100 cm100\text{ cm} behind the lens system.

Step-by-Step Solution

1
Calculate the focal length f1f_1 of the corrective lens required for myopia.
f1=−50 cmf_1 = -50\text{ cm}
The far point distance from the lens is 52 cm−2 cm=50 cm52\text{ cm} - 2\text{ cm} = 50\text{ cm}. A diverging (concave) lens is needed to form a virtual image of distant objects (u=∞u = \infty) at the far point (v=−50 cmv = -50\text{ cm}).
2
Determine the focal length FF of the two lenses in thin contact.
F=+1003 cmF = +\frac{100}{3}\text{ cm}
Using the combined focal length equation 1F=1f1+1f2=−150+120=−2+5100=+3100 cm−1\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = -\frac{1}{50} + \frac{1}{20} = \frac{-2 + 5}{100} = +\frac{3}{100}\text{ cm}^{-1}.
3
Apply the thin lens formula to locate the final image distance vv for u=50 cmu = 50\text{ cm}.
v=+100 cmv = +100\text{ cm}
Using 1F=1u+1v  ⟹  3100=150+1v  ⟹  1v=3100−2100=1100 cm−1\frac{1}{F} = \frac{1}{u} + \frac{1}{v} \implies \frac{3}{100} = \frac{1}{50} + \frac{1}{v} \implies \frac{1}{v} = \frac{3}{100} - \frac{2}{100} = \frac{1}{100}\text{ cm}^{-1}.
4
Determine the nature of the image from the sign of vv.
Real image formed 100 cm100\text{ cm} behind the combined lens system.
A positive value of image distance (v>0v > 0) indicates a real image formed on the opposite side (behind) the lens system.

Key Concept

Thin lens combination and sight defect correction sign conventions
Estimated Time:3m 0s
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Topics

Thin Lenses, Optical Instruments, and Defects of Vision

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