Class 12th

Kindly consider the following figure

In galvanometer

$\Rightarrow \left(l-l_g\right)S=l_{g}G$

$\frac{I_g}{l}=\frac{S}{S+G}$

$\Rightarrow \frac{1}{3}=\frac{S}{S+G}\Rightarrow S+G=3S\Rightarrow G=2S$

${\mu }_B=\frac{C}{v_B}$

$v_B=\frac{3*10^8}{1.47}=20.408*10^7$

$v_A=2.6*10^7+20.408*10^7=23*10^7$

${\mu }_B=\frac{C}{v_B}\&{\mu }_A=\frac{C}{v_A}$

= 1.127

$\approx 1.13$

Optical fibre frequency range is 1 THz to 1000 THz.

Very small change in minority charge carriers produces high value of reverse bias current.

  $v\propto \frac{z}{n}$ $\frac{}{}$

$\frac{}{}$  $v=k\frac{z}{n}$ [K constant]

for 3rd orbit of He+

${}_{}$ $v_{H{e}^{+}}=\frac{k*2}{3}=\frac{2k}{3}$  - (1)

$$ $v_H=\frac{k*1}{3}=\frac{k}{3}$  - (2)

$\frac{v_{H{e}^{+}}}{v_H}=2:1$

Y ->And gate

Velocity of light in medium 2

$=\frac{1}{\sqrt[]{{\mu }_m{\in }_m}}$

$v_2=\frac{1}{\sqrt[]{{\mu }_0{\mu }_r{\in }_0{\in }_r}}$

$=\frac{1}{\sqrt[]{1*{\mu }_0{\in }_0*4}}$

$v_2=\frac{1}{2\sqrt[]{{\mu }_0{\in }_0}}$

By snell's law for total internal Reflection

${\mu }_{2}sin{\theta }_C>{\mu }_{1}sin90°$

${\theta }_c>30°$

$F_R=\left\{\frac{kQ{q}_0}{{\left(a-x\right)}^2}-\frac{kQ{q}_0}{{\left(a+x\right)}^2}\right\}$

$=-kQ{q}_0\left\{\frac{a^2+x^2+2ax-a^2-x^2+2ax}{\left(a^2-x^2\right)}\right\}$

$F_R=-\frac{kQ{q}_0\left(4ax\right)}{{\left(a^2-x^2\right)}^2}$

$T=2\pi \sqrt[]{\frac{a^{3}m}{4kQ{q}_0}=\sqrt[]{\frac{4{\pi }^2{a}^{3}m*4\pi {\epsilon }_0}{4Q{q}_0}}}=\sqrt[]{\frac{4{\pi }^3{\epsilon }_{0}m{a}^3}{Q{q}_0}}$ .                          

Given, l₁ = 1

l₂ = 9l

case 1  $\theta =\frac{\pi }{2}$

$l_P=l_1+l_2+2\sqrt[]{l_1{l}_2}cos\theta$                

= 10 l

case 2 θ = π

$l_Q=l_1+l_2+2\sqrt[]{l_1{l}_2}cos\theta$

$=10l-6l$

$l_P-I_Q=10l-4l=6l$