Physics Ray Optics and Optical Instruments

r + r' = 90°

=> r' = 90 – r = 90 – i

Snell's law :

$\mu sini=1sinr\text{'}$

$\mu sini=sin\left(90-i\right)$           

=> $\Rightarrow sin{i}_c=\frac{1}{\mu }=tani=tanr$

$i_c=si{n}^{-1}\left(tanr\right)$

Since direction of incident ray (DIR) is from left to write, so considering refraction at point M, we can write

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$              

-> $\frac{1.4}{v}-\frac{1.25}{-40}=\frac{1.4-1.25}{-25}$

$\Rightarrow -\frac{1.4}{v}=\frac{0.15}{25}+\frac{1.25}{40}=\frac{1.2+6.25}{200}=\frac{7.45}{200}$

$\Rightarrow v=-\frac{200*1.4}{7.45}=-37.58cm$

from given information : I = A = 60°, r = $\frac{A}{2}=30°$ =>  $\mu =\sqrt[]{3}=\frac{c}{v}$

$\Rightarrow v=\frac{c}{\sqrt[]{3}}\Rightarrow time=\frac{\left(\frac{5\sqrt[]{3}}{100}\right)}{\left(\frac{3*10^8}{\sqrt[]{3}}\right)}=5*10^{-10}s$

As observer is at O So height of water observed by observer

  $\Rightarrow \frac{H}{{\mu }_{\omega }}=\frac{H}{\left(4/3\right)}=\frac{3H}{4}$

given diagram (17.5 - H) is height of observer

$So\frac{3H}{4}=17.5-H$          

$\frac{7H}{4}=17.5$            

7H = 70

H = 10

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

$v=\frac{uf}{u+f}\Rightarrow v=\frac{-f*-f}{-f-f}=\frac{f^2}{-2f}=\frac{-f}{2}$

 

$\therefore$ dist of image from optical cnetre = $\frac{f}{2}$

dist l1 l2 $=2\text{exception 25:}a$

= 2 * 2.3 a

= 4.6 a

$\frac{1}{f_1}=\left({\mu }_1-1\right)\left(\frac{1}{\infty }-\frac{1}{\left(-R\right)}\right)=\frac{{\mu }_1-1}{R}$

$\frac{1}{f_2}=\left({\mu }_2-1\right)\left[\frac{1}{\left(-R\right)}-\frac{1}{\infty }\right]=-\frac{{\mu }_2-1}{R}$

$\frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{{\mu }_1-{\mu }_2}{R}$

$\Rightarrow \frac{R}{f_{eq}}={\mu }_1-{\mu }_2$

$\frac{1}{V_1}+\frac{1}{30}=\frac{1}{10}$

V₁ = 15 cm

$\frac{1}{V_2}-\frac{1}{10}=-\frac{1}{10}$

$\Rightarrow V_2=\infty$

$\Rightarrow V_3=30cm$

$\Rightarrow O{V}_3=75cm$

As image coincides with the object, image formed by convex lens should be at the centre of curvature of the convex mirror. If the convex mirror is removed, then, also, image formed the convex lens is at the same location.

Distance of image from object = 12 + 8 + 2 * 15 = 50 cm

For mirror

${\overrightarrow{V}}_{l_{1}M}=-m^2{\overrightarrow{V}}_{O_{1}M}$

$$ ${\overrightarrow{V}}_{l_{1}M}$  Velocity of image w.r.t. mirror

$$ ${\overrightarrow{V}}_{0M}$  Velocity of object w.r.t. mirror

m magnification

$\frac{1}{v}+\frac{1}{-1.9}=\frac{1}{-0.1}$

$\frac{1}{v}=\frac{1}{1.9}-\frac{1}{0.1}$

$=\frac{0.1-1.9}{1.9*0.1}$

$v=\frac{1.9*0.1}{-1.8}=\frac{0.19}{-1.8}=-0.105$

$m=-\frac{v}{u}=-\left(\frac{-0.1}{-1.9}\right)$

$\left|{\overrightarrow{V}}_{l_{1}M}\right|={\left(\frac{1}{19}\right)}^2\left|{\overrightarrow{V}}_{O_{1}M}\right|$

$=\frac{1}{{\left(19\right)}^2}*40=0.1m/s$