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JEE Main 2014 (11 Apr, Shift 1) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from JEE Main 2014 (11 Apr, Shift 1)

Q97JEE Main 2014 · 11 APR · SHIFT 1Medium

Spherical Surfaces & Thin Lenses

An object is located in a fixed position in front of a screen, Sharp image is obtained on the screen for two positions of a thin lens separated by 10 cm. The size of the images in two situations are in the ratio 3: 2. What is the distance between the screen and the object?

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Answer: (d)

  1. Two lens positions swap uu and vv, so I1I2=(vu)2=32\frac{I_1}{I_2} = \left(\frac{v}{u}\right)^2 = \frac{3}{2}.
  2. vu=1.5≈1.225\frac{v}{u} = \sqrt{1.5} \approx 1.225 and v−u=10v - u = 10 cm.
  3. u=100.225≈44.5u = \frac{10}{0.225} \approx 44.5 cm, v≈54.5v \approx 54.5 cm
  4. D=u+v≈99.0D = u + v \approx 99.0 cm. So the answer is (d).
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