Search any question & find its solution
Question:
Answered & Verified by Expert
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 $\mathrm{cm}$. The size of the images in two situations are in the ratio $3: 3$. What is the distance between the screen and the object?
Options:
Solution:
1392 Upvotes
Verified Answer
The correct answer is:
$99.0 \mathrm{~cm}$
$99.0 \mathrm{~cm}$
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,
$$
\begin{aligned}
&\frac{I_1}{I_2}=\frac{(D+d)^2}{(D-d)^2} \\
&\Rightarrow \quad \frac{3}{2}=\frac{(D+10)^2}{(D-10)^2} \\
&\Rightarrow \quad \frac{3}{2}=\frac{D^2+100+20 D}{D^2+100-20 D} \\
&\Rightarrow 3 D^2+300-60 D=2 D^2+200+40 D \\
&\Rightarrow D^2-100 D+100=0
\end{aligned}
$$
On solving, we get $D=99 \mathrm{~cm}$
Hence the distance between the screen and the object is $99 \mathrm{~cm}$.
$$
\frac{I_1}{I_2}=\frac{3}{2}
$$
Distance of object from the screen, $D=$ ? Applying formula,
$$
\begin{aligned}
&\frac{I_1}{I_2}=\frac{(D+d)^2}{(D-d)^2} \\
&\Rightarrow \quad \frac{3}{2}=\frac{(D+10)^2}{(D-10)^2} \\
&\Rightarrow \quad \frac{3}{2}=\frac{D^2+100+20 D}{D^2+100-20 D} \\
&\Rightarrow 3 D^2+300-60 D=2 D^2+200+40 D \\
&\Rightarrow D^2-100 D+100=0
\end{aligned}
$$
On solving, we get $D=99 \mathrm{~cm}$
Hence the distance between the screen and the object is $99 \mathrm{~cm}$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.