Case $(i):$ when it is a concave mirror of focal length $28 \mathrm{cm}$
$f_{m}=\infty$
Hence $\frac{1}{28}=\frac{2}{f_{1}}+\frac{1}{f_{m}}$
$\Longrightarrow f_{1}=56 \ldots(i)$
Case $(ii):$ when it is a concave mirror of focal length $10 \mathrm{cm}$
Hence $\frac{1}{10}=\frac{2}{f_{1}}+\frac{1}{f_{m}}$
Substituting the value of $f_{1}$, we get
$\Longrightarrow \frac{1}{10}=\frac{2}{56}+\frac{1}{f_{m}}$
$\Longrightarrow \frac{1}{f_{m}}=\frac{1}{10}-\frac{1}{28}$
$\Longrightarrow \frac{1}{f_{m}}=\frac{28-10}{280}$
$\Longrightarrow \frac{1}{f_{m}}=\frac{28-10}{280}$
$\Longrightarrow f_{m}=\frac{280}{18}=\frac{140}{9} \mathrm{cm}$
Therefore, the radius of curvature, $R=2 f_{m}=2 \times \frac{140}{9}=\frac{280}{9} \mathrm{cm}$
Now,
$\frac{1}{f_{1}}=(\mu-1) \frac{1}{R}$
Substituting the respective values, we get
$\frac{1}{56}=(\mu-1) \frac{1}{\frac{280}{9}}$
$\Longrightarrow \mu-1=\frac{280}{9} \times \frac{1}{56}$
$\Longrightarrow \mu=1+\frac{5}{9}=\frac{14}{9}$
is the refractive index of the material of the lens.
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(Avogadro's number $N = 6.02 \times {10^{26}}$atom/kilomol)
$V(z)\, = \,30 - 5{z^2}for\,\left| z \right| \le 1\,m$
$V(z)\, = \,35 - 10\,\left| z \right|for\,\left| z \right| \ge 1\,m$
$V(z)$ does not depend on $x$ and $y.$ If this potential is generated by a constant charge per unit volume $\rho _0$ (in units of $\varepsilon _0$ ) which is spread over a certain region, then choose the correct statement
Major product $(Q)$ of following sequence is
Statement $I$ $m_1$ ,$m_2$ ,$m_3$ remain stationary.
Statement $II$ The value of acceleration of all the $4$ blocks can be determined.
Statement $III$ Only $m_4$ remains stationary.
Statement $IV$ Only $m_4$ accelerates.
Statement $V$ All the four blocks remain stationary.
Now, choose the correct option.
