Physics M.C.Q (1 Marks)

$r_1  =90^{\circ}-c$           $.......(1)$

$\sin c=\frac{1}{\mu}  \Rightarrow \cos c=\frac{\sqrt{\mu^2-1}}{\mu}$

$\Rightarrow$ Apply Snell's law, on incidence surface

$1 \cdot \sin 30^{\circ}=\mu \sin \left(r_1\right) \Rightarrow 1 \times \frac{1}{2}=\mu \times \sin \left(90^{\circ}-c\right)$

$\frac{1}{2}=\mu \times \frac{\sqrt{\mu^2-1}}{\mu}$

On squaring $\frac{1}{4}=\mu^2-1$

$\Rightarrow \mu^2=\frac{5}{4} \Rightarrow \mu=\frac{\sqrt{5}}{2}$

For distant object,

$m=\frac{f_0}{f_e}=\frac{140}{5}=28$

speed of light is a medium $V_2=\frac{10 x}{t_2}$

$\sin \theta_c=\frac{V_2}{V_1}=\frac{10 x}{t_2} \frac{t_1}{x}$

$\theta_c=\sin ^{-1}\left(\frac{10 t_1}{t_2}\right)$

$\frac{1}{f_1}=[1.6-1]\left[\frac{1}{\infty}-\frac{1}{20}\right]=\frac{-3}{100}$

$\frac{1}{f_2}=[1.5-1]\left[\frac{1}{20}+\frac{1}{20}\right]=\frac{1}{20}$

$\frac{1}{f_3}=\frac{-3}{100}$

$\frac{1}{f_{\text {eq }}}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}$

$\frac{1}{f_{\text {eq }}}=-\frac{3}{100}+\frac{1}{20}-\frac{3}{100}=\frac{-1}{100}$

$\frac{1}{f_{e q}}=\frac{1}{f}-\frac{1}{f}$

$f_{e q}=\infty$

By Snell's law

$1 \sin 60^{\circ}=\sqrt{3} \sin r$

$\frac{\sqrt{3}}{2}=\sqrt{3} \sin r$

$\sin r=\frac{1}{2}$

$r=30^{\circ}$

Angle between refracted and reflected ray is $90^{\circ}$

Method $(ii)$

Because angle of incidence is Brewster's angle so that angle between reflected and refracted ray is $90^{\circ}$

$\operatorname{tani}_{p}=\mu=\sqrt{3}$

$i_{p}=60^{\circ}= i$

Critical angle

$\text { Sini i }_{c}=\frac{\mu_{ R }}{\mu_{ D }}=\frac{ u _{ D }}{ u _{ R }}=\frac{1.5}{2}=\frac{3}{4}$

$i _{c}=\sin ^{-1}\left(\frac{3}{4}\right)$

$\sin i_{c}=\frac{\mu_{R}}{\mu_{D}}=\frac{u_{D}}{u_{R}}$

$i_{c}=\sin ^{-1}\left(\frac{3}{4}\right)$

$\mu=\frac{3}{2}$

$P =\frac{1}{ f }=(\mu-1)\left(\frac{1}{ R _{1}}-\frac{1}{ R _{2}}\right)$

$P=\left(\frac{3}{2}-1\right)\left(\frac{1}{0.2}+\frac{1}{0.2}\right)$

$P=\frac{1}{2}\left(\frac{2}{0.2}\right)=\frac{10}{2}=+5 D$

$f _e=2\,cm$

for Normal adjustment

$\rightarrow \text { M.P. }=\frac{f_0}{f_e}=\frac{2000}{2}=1000$

$\rightarrow$ Distance between both lens $= f _0+ f _e$

$=2000+2$

$=2002\,cm$

$=20.02\,m$

$\rightarrow$ Image is inverted and magnified

$\rightarrow$ Aperture of eye piece is smaller than objective.

$=20-5$

$=15\, \mathrm{~cm}$

$r_{2}=30^{\circ}\left(r_{1}=0^{\circ}\right)$

from Snell's law

$\sqrt{3} \sin r_{2}=1 \times \sin e$

$\text { R.P. }=\frac{\mathrm{a}}{1.22 \lambda}$

Large aperture$(a)$ of the objective lens provides bettern resolution

$\therefore$ good quality of image is formed and also it gathers more light.

$\mathrm{u}=-60\, \mathrm{~cm}$

$\mathrm{f}=+30\, \mathrm{~cm}$

$\Rightarrow \mathrm{v}=\frac{\mathrm{uf}}{\mathrm{u}+\mathrm{f}}=\frac{-60 \times 30}{-60+30}=60\, \mathrm{~cm}$

This real image formed by lens acts as virtual object for mirror

Real image from plane mirror is formed $20 \mathrm{~cm}$ in front of mirror, hence at $20 \mathrm{~cm}$ distance from lens. Now for second refraction from lens,

$\mathrm{u}=-20\, \mathrm{~cm}$

$\mathrm{f}=+30\, \mathrm{~cm}$

$v=\frac{\mathrm{uf}}{\mathrm{u}+\mathrm{f}}=\frac{-20 \times 30}{-20+30}=-60\, \mathrm{~cm}$

So, final virtual image is $60\, \mathrm{~cm}$ from lens, or $20\, \mathrm{~cm}$ behind mirror

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

$\frac{1}{v}-\frac{1}{\frac{3 f}{2}}=\frac{1}{-f}$

$\frac{1}{v}-\frac{2}{3 f}=-\frac{1}{f}$

$\frac{1}{v}=-\frac{1}{f}+\frac{2}{3 f}$

$\frac{1}{v}=\frac{-3+2}{3 f}$

$v=-3 f$

$\sin 45^{\circ}=\frac{1}{\mu}$

$\frac{1}{\sqrt{2}}=\frac{1}{\mu}$

$\mu=\sqrt{2}$

Velocity of light in medium

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

$v =\frac{3 \times 10^{8}}{\sqrt{2}} m / s$

as $P=\frac{1}{f(\text { in } m)}$

$10=(\mu-1)\left(\frac{1}{0.10}+\frac{1}{0.10}\right)$

$(\mu-1)=\frac{1}{2}$

$\mu=\frac{3}{2}$

$r_{1}=A$

Apply Snell's law

$\sin 1=\mu \sin r _{1}$

for small angle $\left( r _{1}= A \right)$

$i=\mu A$

$\Rightarrow \frac{1}{25}=\left(\frac{1.5-1}{1}\right)\left(\frac{1}{R}-\frac{1}{(-2 R)}\right) $

$\Rightarrow \frac{2}{25}=\frac{3}{2 R}$

$\Rightarrow R=\frac{75}{4}=18.75$

$\Rightarrow \mathrm{F}_{1}=\mathrm{f} / 2$

$\Rightarrow \mathrm{F}_{2}=\mathrm{f}$

$\Rightarrow \frac{\mathrm{F}_{1}}{\mathrm{F}_{2}}=\frac{1}{2}$

angle of refraction $= 90^o$

$\frac{1}{f}=\frac{1}{v_{1}}+\frac{1}{u_{1}} ;-\frac{1}{15}=\frac{1}{v_{1}}-\frac{1}{40} \Rightarrow \frac{1}{v_{1}}=\frac{1}{-15}+\frac{1}{40}$

$v_{1}=-24 \,\mathrm{cm}$

When object is displaced by $20 \,\mathrm{cm}$ towards mirror Now, $u_{2}=-20\, \mathrm{cm}$

$\frac{1}{f}=\frac{1}{v_{2}}+\frac{1}{u_{2}} ; \frac{1}{-15}=\frac{1}{v_{2}}-\frac{1}{20} \Rightarrow \frac{1}{v_{2}}=\frac{1}{20}-\frac{1}{15}$

$v_{2}=-60\, \mathrm{cm}$

So, the image will be shift away from mirror by $(60-24) \mathrm{cm}=36 \,\mathrm{cm}$

Angular resolution $=\frac{D}{1.22 \lambda}$ should be large.

So, objective lens should have large focal length $\left(f_{o}\right)$ and large diameter $D$ for large angular

Applying Snell's law at point $M,$

$\frac{\sin i}{\sin 30^{\circ}}=\frac{\sqrt{2}}{1} \Rightarrow \sin i=\sqrt{2} \times \frac{1}{2}$

$\sin i=\frac{1}{\sqrt{2}} i . e_{.}, i=45^{\circ}$

Given $\mu=1.42, A=10^{\circ}, \mu^{\prime}=1.7, A^{\prime}=?$

$\therefore \quad(1.42-1) \times 10=(1.7-1) A^{\prime}$

$(0.42) \times 10=0.7 \times A^{\prime}$

or $\quad A^{\prime}=\frac{0.42 \times 10}{0.7}=6^{\circ}$

$v=-(22-2)=-20$

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

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

$f =\frac{1200}{-40}$

$f =30$

$\quad \tan 2 \theta \approx 2 \theta=\frac{y}{x}$

$\therefore \theta=\frac{y}{2 x}$

$r =\frac{ A }{2} r =\frac{60^{\circ}}{2} r =30^{\circ}$

$m=-2 \Rightarrow v=2 u$

As $v$ and $u$ have same signs so the mirror is concave and image formed is real.

$m=-\frac{1}{2} \Rightarrow v=\frac{u}{2} \Rightarrow$ Concave mirror and real image.

$m=+2 \Rightarrow v=-2 u$

As $v$ and $u$ have different signs but magnification is $2$ so the mirror is concave and image formed is virtual.

$m=+\frac{1}{2} \Rightarrow v=-\frac{u}{2}$

As $v$ and $u$ have different signs with magnification $\left(\frac{1}{2}\right)$ so the mirror is convex and image formed is virtual.

 $l=$ length of the slab 

$x=$ position of air bubble from one side As per question, total apparent length of slab $=5+3$

or $\frac{x}{\mu}+\frac{(l-x)}{\mu}=8$ or $\frac{l}{\mu}=8 \therefore l=8 \mu=8 \times 15=12\, \mathrm{cm}$

since the ray undergoes minimum deviation, therefore, angle of emergence from second face, $e=i=45^o$

$\therefore \quad \delta_{m}=i+e-A=45^o+45^o-60^o=30^o$

$\mu  = \frac{{\sin \left( {\frac{{A + {\delta _m}}}{2}} \right)}}{{\sin \left( {\frac{A}{2}} \right)}}$ $ = \frac{{\sin \left( {\frac{{{{60}^o} + {{30}^o}}}{2}} \right)}}{{\sin \left( {\frac{{{{60}^o}}}{2}} \right)}}{\text{ }}$

$ = \frac{{\sin {{45}^o}}}{{\sin {{30}^o}}} = \frac{1}{{\sqrt 2 }} \times \frac{2}{1} = \sqrt 2 $

Using lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

or $\frac{1}{\infty}-\frac{1}{4}=\frac{1}{f}$ or $f=-4\, \mathrm{m}$

Lens should be concave.

Power of lens $=\frac{1}{f}=\frac{1}{-4}=-0.25\, \mathrm{D}$

Tube length( $(l)=$ Distance between lenses $=v_{o}+f_{e}$ For objective lens,

${u_o} =  - 200{\text{cm}},{v_o} = ?$

$\frac{1}{v} - \frac{1}{{{u_o}}} = \frac{1}{{{f_o}}}$  or   $\frac{1}{{{v_o}}} - \frac{1}{{ - 200}} = \frac{1}{{40}}$

or $\frac{1}{{{v_0}}} = \frac{1}{{40}} - \frac{1}{{200}} = \frac{4}{{200}}$  $\therefore {v_0} = 50\,{\text{cm}}$

$\therefore \quad l=50+4=54\, \mathrm{cm}$

Focal length $(f)$ of glass convex lens is given by

$\frac{1}{f}=\left(\mu_{g}-1\right)\left(\frac{2}{R}\right)$

or $\frac{1}{f}=\left(\frac{3}{2}-1\right) \frac{2}{R}=\frac{1}{R}$ or $f=R.........(i)$

Focal length $(f)$ of water filled concave lens is given by

$\frac{1}{f^{\prime}} =\left(\mu_{w}-1\right)\left(-\frac{2}{R}\right) \text { or } \frac{1}{f^{\prime}}=\left(\frac{4}{3}-1\right)\left(-\frac{2}{R}\right) $

$=-\frac{2}{3 R}=-\frac{2}{3 f} \quad[\text { Using } \operatorname{eqn} .(\mathrm{i})]$

Equivalent focal length $\left(f_{e q}\right)$ of lens system $\frac{1}{f_{e q}}=\frac{1}{f}-\frac{2}{3 f}+\frac{1}{f}=\frac{3-2+3}{3 f}=\frac{4}{3 f}$

$\therefore f_{e q}=\frac{3 f}{4}$

$ = \frac{1}{2}(10 + 170) = 90\,cm$

and so will form $\left({\frac{{360}}{{90}} - 1} \right) = $$4 -1 $ i.e. $3$ images of the person.

Now these images with person will act as objects for the ceiling mirror and so ceiling mirror will form $4$ images further. 

Therefore total number of images formed $ = 3 + 3 + 1 = 7$ 

Note : He can see. $6$ images of himself.

So in order to see a man of height h, the mirror should be of height $\frac{h}{2}$

In the ceiling mirror the original velocity will be seen.

$ = \sqrt {{{(6)}^2} + {{(1.5)}^2}} = 6.18\,m$
 

Hence velocity of approach of image towards man is $2(u -v).$

$=100-(-100)$

$=200 cm$

Now, object approaches the mirror with the speedv $=10 cm / s$ So, distance travelled by object in $6\; sec$

$D =10 \times 6$

$D =60 cm$

At this instant,

Distance between mirror and object

$D =100-60$

$D =40 cm$

Now, the distance between image and object is

$D =40-(-40)$

$D =80\;0 cm$