Search any question & find its solution
Question:
Answered & Verified by Expert
An object placed in front of a concave mirror at a distance of $x$ cm from the pole gives a 3 times magnified real image. If it is moved to a distance of $(x+5) \mathrm{cm},$ the magnification of the image becomes 2 . The focal length of the mirror is
Options:
Solution:
1338 Upvotes
Verified Answer
The correct answer is:
$30 \mathrm{cm}$
Given $y_{1}=3 x$ and $v_{2}=2(x+5)$ So here, Ist case
$$
\begin{array}{r}
\frac{1}{-x}+\frac{1}{-3 x}=\frac{1}{f} \\
\frac{1}{-(x+5)}+\frac{1}{-2(x+5)}=\frac{1}{f}
\end{array}
$$
On solving Eqs. (i) and (ii) we get $f=30 \mathrm{cm}$
$$
\begin{array}{r}
\frac{1}{-x}+\frac{1}{-3 x}=\frac{1}{f} \\
\frac{1}{-(x+5)}+\frac{1}{-2(x+5)}=\frac{1}{f}
\end{array}
$$
On solving Eqs. (i) and (ii) we get $f=30 \mathrm{cm}$
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.