Ray Optics

JEE Main level, single correct, medium

287 of 295

Question 287 of 295

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\mathrm{cm}. The size of the images in two situations are in the ratio 3:33: 3. What is the distance between the screen and the object?

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Answer: (D) 99.0 cm99.0 \mathrm{~cm}

Given: Separation of lens for two of its position, d=10 cmd=10 \mathrm{~cm} Ratio of size of the images in two positions
I1I2=32\frac{I_1}{I_2}=\frac{3}{2}
Distance of object from the screen, D=D= ? Applying formula,
I1I2=(D+d)2(D−d)2⇒32=(D+10)2(D−10)2⇒32=D2+100+20DD2+100−20D⇒3D2+300−60D=2D2+200+40D⇒D2−100D+100=0\begin{aligned} &\frac{I_1}{I_2}=\frac{(D+d)^2}{(D-d)^2} \\ &\Rightarrow \quad \frac{3}{2}=\frac{(D+10)^2}{(D-10)^2} \\ &\Rightarrow \quad \frac{3}{2}=\frac{D^2+100+20 D}{D^2+100-20 D} \\ &\Rightarrow 3 D^2+300-60 D=2 D^2+200+40 D \\ &\Rightarrow D^2-100 D+100=0 \end{aligned}
On solving, we get D=99 cmD=99 \mathrm{~cm}
Hence the distance between the screen and the object is 99 cm99 \mathrm{~cm}.

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