Physics M.C.Q (1 Marks)

$ \Rightarrow \frac{I}{{ + 6}} = \frac{{ - f}}{{ - f - ( - 4f)}}$$ \Rightarrow I = - \,2\,cm.$

$ = 30' = {\left( {\frac{1}{2}} \right)^o} = \frac{\pi }{{360}}\,\,rad$

If $x$ be the diameter of the image of the sun, then

$\frac{{{\rm{Arc}}}}{{{\rm{Radius}}}} = \frac{x}{{100}} = \frac{1}{2}.\frac{{2\pi }}{{360}} = \frac{\pi }{{360}}$

$ \Rightarrow \,\,x = \frac{{100\pi }}{{360}} = 0.87\,cm$

According to property of plane mirror it will form image at a distance of $30 cm$ behind it. Hence for convex mirror $u = -50 cm, \,\,v = + 10 cm$ 

By using $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ 

$ \Rightarrow \,\,\,\frac{1}{f} = \frac{1}{{ + 10}} + \frac{1}{{ - 50}} = \frac{4}{{50}}$ 

$ \Rightarrow $ $f = \frac{{25}}{2}cm$ 

$ \Rightarrow \,\,\,R = 2f = 25cm.$

For real image $m$ $ = - \frac{f}{{(u - f)}}$$ = - \frac{f}{x}$

$ = - \frac{f}{{{\rm{(Distance\, of \,object\, from\, focus}})}}$

==> $m \propto \frac{1}{x}$.

$\frac{1}{ v }+\frac{1}{ u }=\frac{1}{ f }$

$\frac{1}{ v }-\frac{2}{15}=\frac{-1}{5}$

$v =-15$

i.e, image is formed at pole of convex mirror for convex mirror, $u=0 ; f=+5$

$\frac{1}{ v }+\frac{1}{ u }=\frac{1}{ f }$

$\frac{1}{ v }+\frac{1}{0}=\frac{1}{5}$

$v =0$

so final image is frormed at pole of convex mirror

Object length $h=4$

Real image length $h ^{\prime}=1 cm$

We know that,

$m =-\frac{ v }{ u }=\frac{- f }{- f -( u )}$

$-\frac{1}{ u }=\frac{-12}{-12- u }$

$12+ u =-48$

$u =-48-12$

$u =-60 cm$

Hence, the object distance is $60\; cm$

$\frac{1}{v}+\frac{1}{\infty}=\frac{1}{f}$

$v=f$

image is formed at the focus $F$ of concave mirror. If $A^{\prime} B^{\prime}$ is the size/diameter of the image of the sun, then from

$\theta=\frac{A^{\prime} B^{\prime}}{P F}=\frac{d}{f}$

$d=\theta f$

$\Rightarrow- 3=\frac{{ - 30}}{{ - 30 - u}}$

$\Rightarrow u= - 40\,cm$

Differentiating both sides $-\frac{1}{v^2} \frac{d v}{d t}=\frac{1}{u^2} \frac{d u}{d t}$

$\frac{d v}{d t}=v_l=-\left(\frac{v}{u}\right)^2 \frac{d u}{d t}=-\left(\frac{v}{u}\right)^2 v_0$

Again $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{2}{r}-\frac{1}{u}=\frac{2 u-r}{r u}$

$v=\frac{u r}{2 u-r}$

$v_i=-\left(\frac{v}{u}\right)^2 v_o=-v_o\left(\frac{r}{2u-r}\right)^2$

$\mathrm{v}=3 \mathrm{x}$

$3 x+x+80$

$4 x=80 \Rightarrow x=20$

$\mathrm{u}=-20 \quad \mathrm{v}=60$

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\frac{1}{60}-\frac{1}{20}=\frac{1}{f}$

$f=-30 \mathrm{\,cm}$

$ \Rightarrow \,\,\,x = - 4.5m\,\, = \,\, - 450cm.$

$1.8 f+18=f$

$18=-0.8 f$

$f=-22.5 \mathrm{\,cm}$

Tn second case, $u=-50$

$\Rightarrow$ object is beyond $C.$

$\Rightarrow$ image is inverted and diminished.

= $|m|$ (object speed)

= $4\,\, mm/s$

$\left. \begin{array}{l}
R.O. \to V.I.\\
V.O. \to R.I.
\end{array} \right\}always\,\,erect$

$\mathrm{u}=\mathrm{f}\left(1-\frac{1}{\mathrm{m}}\right)=17.5\left(1-\frac{1}{2.5}\right)=10.5 \mathrm{\,cm}$

(magnification) $m=\frac{f}{f-u}$

Focal real image $m=-n$

$-n=\frac{-f}{-f-u}$

$\Rightarrow u=-\frac{f(n+1)}{u}$

$f=\frac{R}{2}=5 \,cm$

$\theta=\frac{10}{5}=2 \,rad$

$\frac{1}{v} - \frac{1}{u}=\frac{1}{{{f}}}$

$\Rightarrow \,\,\frac{1}{v}-\frac{1}{{( - 15)}}=\frac{1}{{10}}$

$\Rightarrow v=30\,cm$

$x=v+2 f \Rightarrow 30+3 \times 12=54 \;cm$

$\left(V_{I / M}\right)=-m_{T}^{2} .\left(V_{O / M}\right)_{\mathrm{M}}$

so image $I$ in concave mirror will move towards right.

$\mathrm{f}=+5 \mathrm{\,cm}$

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$\mathrm{v}=+4 \mathrm{\,cm}$

Distance between object and image is $24 \mathrm{\,cm} .$ So plane mirror must be placed in between i.e. at $12 \mathrm{\,cm}$ from object.

so that bottom of the container should appear to be raised upto height $(21-x).$

As shown in figure apparent depth $h' = (21 - x)$
Real depth $h = x$

$\therefore $$\mu = \frac{h}{{h'}} \Rightarrow \frac{4}{3} = \frac{x}{{21 - x}} \Rightarrow x = 12 cm$

= $\frac{{H/4}}{{{\mu _1}}} + \frac{{H/4}}{{{\mu _2}}} + \frac{{H/4}}{{{\mu _3}}} + \frac{{H/4}}{{{\mu _4}}}$

$ = \frac{H}{4}\left( {\frac{1}{{{\mu _1}}} + \frac{1}{{{\mu _2}}} + \frac{1}{{{\mu _3}}} + \frac{1}{{{\mu _4}}}} \right)$

$= 2\,mm.$

$ = 12 \times \left( {1 - \frac{3}{4}} \right)$$= 3\, cm.$

==> $\frac{1}{{\sqrt 3 }} = \frac{{\sin r}}{{\sin i}} = \frac{1}{\mu }$ ==> $\mu = \sqrt 3 $

Also $v = \frac{c}{\mu } = nc$

==> $n = \frac{1}{\mu } = \frac{1}{{\sqrt 3 }} = {(3)^{ - 1/2}}$