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?......\(cm\)
Pick an option to check your answer.
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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