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JEE Mains · Physics · STD 12 - Ray optics and optical instruments
  • JEE Mains 2014
  • Hard
  • Numerical

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?......\(cm\)

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Answer & Solution

Correct Answer

(D) \(99\)

Step-by-step Solution

Detailed explanation

Given: Separation of lens for two of its position, \(d=10 \mathrm{cm}\)

Ratio of size of the images in two positions

\(\frac{I_{1}}{I_{2}}=\frac{3}{2}\)

Distance of object from the screen, \(D=?\)

Applying formula,

\({\frac{{{I_1}}}{{{I_2}}} = \frac{{{{(D + d)}^2}}}{{{{(D - d)}^2}}}}\)

\({ \Rightarrow \,\,\frac{3}{2} = \frac{{{{(D + 10)}^2}}}{{{{(D - 10)}^2}}}}\)

\({ \Rightarrow \quad \frac{3}{2} = \frac{{{D^2} + 100 + 20D}}{{{D^2} + 100 - 20D}}}\)

\(\Rightarrow 3 D^{2}+300-60 D=2 D^{2}+200+40 D\)

\(\Rightarrow \quad D^{2}-100 D+100=0\)

On solving, we get \(D=99\, \mathrm{cm}\)

Hence the distance between the screen and the object is \(99\, \mathrm{cm} .\)

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