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Q.Two lenses of power +15D and −5D are in contact with each other forming a combination lens.

i) What is the focal length of this combination?
ii) An object of size 3cm is placed at 30cm from this combination of lenses. Calculate the position and size of the image formed. OR Define power of accommodation of human eye. What causes myopia eye? How can it be corrected?
Nagaland NbseNagaland Board of School Education 2022Subjective· 3mImportance★★★★★
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Powers in contact simply add; use the lens formula and magnification relation to locate and size the image.

  1. Focal length of the combination: For thin lenses in contact, the net power is the algebraic sum of the individual powers: P=P1+P2=(+15 D)+(−5 D)=+10 DP = P_1 + P_2 = (+15\,D) + (-5\,D) = +10\,D Since P=1f (in metres)P = \dfrac{1}{f\,(\text{in metres})}: f=1P=110 m=10 cm  (converging, since P>0)f = \frac{1}{P} = \frac{1}{10}\ \text{m} = 10\ \text{cm} \ \ (\text{converging, since } P>0)
  2. Position and size of the image: Using the Cartesian sign convention, object distance u=−30 cmu = -30\ \text{cm}, focal length f=+10 cmf = +10\ \text{cm}. From the lens formula: 1v−1u=1f  ⟹  1v=1f+1u=110+(1−30)=3−130=230=115\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \left(\frac{1}{-30}\right) = \frac{3-1}{30} = \frac{2}{30} = \frac{1}{15} v=+15 cmv = +15\ \text{cm} The positive vv means the image forms 15 cm on the far side of the lens combination (a real image). Magnification: …

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