Physics M.C.Q (1 Marks)

$⇒ i = 30°$

$\tan \theta = \frac{x}{d}$

$ \Rightarrow \,\,\,\,\,\,n = \frac{l}{{d\tan \theta }}$

$ \Rightarrow \,\,\,\theta = {30^o}$.

પ્રતિબિંબની સંખ્યા $ = \frac{{360}}{{30}} - 1 = 11$

$v_{\text {image1 }}=v_{\text {image1, mirror } 1}+V_{\text {mirror } 1, \mathrm{obj}}$

$=v+v=2 v$

Now this image becomes object for mirror $2 .$ With respect to mirror $2$ the image is going away towards left with speed $3v.$ So its second image will more away towards right with speed $3v$ w.r.t mirror $2.$ Hence speed w.r.t $O$ will be $3v + v =$$4 \mathrm{v}$

Hence for nth image $v_{\text {image }}=2 n v$

The second nearest image in the right mirror forms with the nearest image of the man in the left mirror as the object as shown in the figure.

So the distance of the second nearest image from the man is:

$1+5=6 m$

where  ${\bar V_I}=$ velocity of image

             ${\bar V_0}=$ velocity of object

             ${\bar V_m}=$ velocity of mirror

$v_{0} =40 \mathrm{\,cm} / \mathrm{s} $

w.r.t. mirror

${\overrightarrow {\rm{v}} _{{\rm{Im}}}} =  - 3\widehat {\rm{i}} + 5\sqrt {3\widehat {\rm{j}}} $

$78\,cm$ from ${M_1}$

$\frac{2 \mathrm{L}}{3}+\mathrm{L}+\frac{\mathrm{L}}{3}=2 \mathrm{L}$

surface between ${2^{nd}}$ inte

$\frac{4 \mathrm{L}}{3}+\mathrm{L}+\frac{5 \mathrm{L}}{3}=4 \mathrm{L}$

$\frac{\mathrm{x}}{\mathrm{d}} =\tan 53^{\circ} $

$\Rightarrow \quad \mathrm{x} =\mathrm{d} \tan 53^{\circ} $

$=10 \times \frac{4}{3} \mathrm{\,cm}$

Total no. of reflections

$=\frac{20}{10 \times \frac{4}{3}} \times 100$ $=150$

${\overrightarrow v _{rel}} = {\overrightarrow v _{in}} - {\overrightarrow v _{obj}} = 3\hat i - 4\hat j - 3\hat i - 4\hat j =  - 8\hat j$

$\sqrt {2gl}  + \sqrt {2gl}  = 2\sqrt {2gl} $

$\vec{V}_{l_0}=\vec{V}_l-\bar{V}_0$

$\vec{V}_{l_0}=-u \cos \theta \hat{i}+u \sin \hat{j}-(u \cos \theta \hat{i}+u \sin \theta \hat{j})$

$\vec{V}_{l_0}=-2 u \cos \theta \hat{i}$

Straight line and horizontal

So spot on the screen will make $2n$ revolution per second.

$=180^{\circ}-2 \mathrm{i}$

$=180^{\circ}-2\left(90^{\circ}-\theta\right)$

$=2 \theta$

Angle between the mirror $\theta=45$

no. of Imager $=\frac{360}{45}-1=7$

By using mirror formula $\frac{1}{f} = \frac{1}{{ - \frac{u}{n}}} + \frac{1}{u}$

$ \Rightarrow u = - (n - 1)f$

$Plane\, Mirrors$ always produce a virtual and erect image having the same size as that of the object irrespective of the position of object.

$Concave\, mirrors$ produce real and virtual, erect and inverted, diminished, samesized and magnified image depending upon the position of the object on the principal axis. The concave mirror forms a virtual and erect image only when the object is placed between the Focus and the pole of the mirror.

Thus, a virtual image can be formed by all the three mirrors.

Hence, the correct answer is OPTION $D.$

$ \Rightarrow \frac{I}{{ + (7.5)}} = \frac{{(25/2)}}{{\left( {\frac{{25}}{2}} \right) - ( - 40)}} \Rightarrow I = 1.78\;cm$

The focal length $f = \frac{{uv}}{{u + v}} = \frac{{(f + {x_1})(f + {x_2})}}{{(f + {x_1}) + (f + {x_2})}}$

On solving, we get ${f^2} = {x_1}{x_2}$ or $f = \sqrt {{x_1}{x_2}} $

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

$ \Rightarrow \frac{{ - v}}{u} = \frac{f}{{f - u}}$ so $m = \frac{f}{{f - u}}$.

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \Rightarrow \frac{1}{f} = \frac{1}{{ - 20}} + \frac{1}{{( - 10)}} \Rightarrow f = \frac{{20}}{3}\,\,cm.$

If object moves towards the mirror by $0.1\,cm$ then.

$u = (10 - 0.1) = 9.9\,cm.$ Hence again from mirror formula

$\frac{1}{{ - 20/3}} = \frac{1}{{v'}} + \frac{1}{{ - 9.9}} \Rightarrow v' = 20.4\,cm$ i.e. image shifts away from the mirror by $0.4\,cm.$

$\because \,m =  - \frac{v}{u} \Rightarrow v =  - 2u$. By using mirror formula

$\frac{1}{{ - 15}} = \frac{1}{{( - 2u)}} + \frac{1}{u} \Rightarrow u = - 7.5\,cm$