Question

Focal length of a concave mirror is $-20cm$. Match the object distance(Real object) given in column $II$ corresponding to magnification (only magnitude) given in column $I$.
$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \left(A\right)2& \left(p\right)10cm\\ \left(B\right)\frac{1}{2}& \left(q\right)30cm\\ \left(C\right)1& \left(r\right)20cm\\ \left(D\right)\frac{1}{4}& \left(s\right)\text{none of these}\end{array}$

• A. $\left(A\right)\to p,\left(B\right)\to s,\left(C\right)\to s,\left(D\right)\to s$
• B. $\left(A\right)\to p,q,\left(B\right)\to r,\left(C\right)\to q,\left(D\right)\to s$
• C. $\left(A\right)\to p,\left(B\right)\to r,\left(C\right)\to s,\left(D\right)\to s$
• D. $\left(A\right)\to p,q,\left(B\right)\to s,\left(C\right)\to s,\left(D\right)\to s$

Answer 1

The correct option is D $\left(A\right)\to p,q,\left(B\right)\to s,\left(C\right)\to s,\left(D\right)\to s$

For concave mirror, $f=-20cm$
$\left(A\right)$: $\text{Image is virtual}$
$m=2=\frac{-v}{u}\Rightarrow v=-2u$
Using, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f},$
we get, $\frac{-1}{2u}+\frac{1}{u}=\frac{-1}{20}$
$\Rightarrow \frac{1}{2u}=\frac{-1}{20}\Rightarrow u=-10cm$
Object is $10cm$ infront of mirror.

If image is real,
$m=-2=\frac{-v}{u}\Rightarrow v=2u$
$\Rightarrow \frac{1}{2u}+\frac{1}{u}=\frac{-1}{20}\Rightarrow u=-30cm$
Object is $30cm$ infront of mirror.
Hence, correct options are : $\left(A\right)\to p,q$

$\left(B\right)Here,m=\frac{1}{2}<1$
Image has to be real( As virtual images in concave mirror are enlarged), magnification of a real image is negative. So,
$\Rightarrow \frac{-1}{2}=\frac{-v}{u}\Rightarrow v=\frac{u}{2}$
Using, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f},$
we get,
$\frac{1}{\frac{u}{2}}+\frac{1}{u}=\frac{-1}{20}\Rightarrow \frac{3}{u}=\frac{-1}{20}\Rightarrow u=-60cm$
Object is $60cm$ infront of mirror.
Hence, correct option is $\left(B\right)\to s$,

$\left(C\right)\text{If}m=1,u=2f\Rightarrow u=-40cm$
Hence correct option is $s$.
$\Rightarrow \left(c\right)\to \left(s\right)$
(D) Here,$m=\frac{1}{4}<1$
Images has to be real( As virtual images in concave mirror are enlarged). It is possible only when object is beyond $C=2f=-40cm$
So, answer is none of these.
$\Rightarrow \left(D\right)\to \left(s\right)$