Class 12th

Kindly consider the following figure

Kindly consider the following figure

Clearly all equations can be $\lambda x-2y-6z=0$  form where $\lambda =1,\frac{4}{3},\frac{6}{5}$  

So for $x=0y=-3z$

So $x^2+y^2+z^2=10z^2$  (as  ) $x=0\mathrm{}y=-3z$

ATQ $10\le 10z^2\le 200$

$1\le z^2\le 20\Rightarrow z=\pm 1,\pm 2,\pm 3,\pm 4$

we know, = $\frac{\lambda }{d}$ $\frac{}{}$ $\mu =\frac{C}{V}=\frac{C}{\upsilon \lambda }$ $\frac{}{}\frac{}{}$

${}_{}\frac{}{{}_{}{}_{}}$  ${\lambda }_1=\frac{C}{{\mu }_1{\upsilon }_1}$ [with change in medium frequency remain constant]

${\lambda }_2=\frac{C}{{\mu }_2{\upsilon }_2}$

$\frac{{\lambda }_2}{{\lambda }_1}=\frac{{\mu }_1}{{\mu }_2}=\frac{5}{7}$ ${\lambda }_2=\frac{5}{7}{\lambda }_1$

${\theta }_1=\frac{{\lambda }_1}{d_1}$

2= $\frac{{\lambda }_2}{d_2}$

$\frac{{\theta }_2}{{\theta }_1}=\frac{{\lambda }_2}{{\lambda }_1}[?d_1=d_2]$

2 $=\frac{5}{7}*{\theta }_1$

${\theta }_2=\frac{5}{7}*0.35$

Equation of tangent of at $y=\mathrm{l}\mathrm{o}\mathrm{g}?x$  at $\left(e^{\lambda },\lambda \right)$  is $y-\lambda =\frac{1}{e^{\lambda }}\left(x-e^{\lambda }\right)$

It cuts y   axis at $(0,\lambda ,-1)$

Also normal to $y^2=8x$    at $\left(\mathrm{7,4}\right)$    is $y-4=-(x-2)$  

  $y=-x+6\mathrm{}$  Cuts y axis at $\left(\mathrm{0,6}\right)\mathrm{}$  SO ATQ $6=-1+\lambda \Rightarrow \lambda =7$

$\frac{B_0^2}{2{\mu }_0}C$

$B_0^2=\frac{I2{\mu }_0}{C}$

$B_0^2=\frac{0.22*2*4*10^{-7}}{9*108}$

B₀ = 42.9 * 10^-9

42.9 Ans

Plane containing both line $x=1$

Image of point $(\mathrm{2,1},3)$  in the plane $\frac{\alpha -2}{1}=\frac{\beta -1}{0}=\frac{\gamma -3}{0}=-2\left(1\right)$

$\alpha =0,\beta =1,\gamma =3;\alpha +\beta +\gamma =4$

$\lambda =\frac{h}{p}or\lambda =\frac{h}{mv}$

${\lambda }_p={\lambda }_{\alpha }\left[Given\right]$

$\frac{h}{m_p{V}_p}=\frac{h}{m_{\alpha }{v}_{\alpha }}\Rightarrow \frac{v_p}{v_{\alpha }}=\frac{m_{\alpha }}{m_p}$

$\Rightarrow \frac{v_p}{v_{\alpha }}=\frac{4m}{m}=\frac{4}{1}$ or 4 : 1

  $x-cy-bz=0$

$cx-y+az=0$

$bx+ay-z=0$

Equation of any plane passing through the line of intersection of planes (1) and (2)

$x-cy-bz+\lambda (cx-y+az)=0$

$(1+\lambda c)x-(c+\lambda )y+(-b+a\lambda )z=0$

If (3) &  (4) is same plane.

$\frac{1+c\lambda }{b}=\frac{-(c+\lambda )}{a}=\frac{-b+a\lambda }{-1}$

(i)

(ii) (iii)

By (i)  & (ii) $\lambda =-\frac{(a+bc)}{ac+b}$

By (ii) & (iii)

$\lambda =\frac{-(ab+c)}{1-a^2};a^2+b^2+c^2+2abc=1$

Amplitude is proportional to the slit – width, thus,

^{$\frac{A_1}{A_2}=3$}

^{$\frac{l_{max}}{l_{min}}=\frac{{\left(A_1+A_2\right)}^2}{{\left(A_1-A_2\right)}^2}=\frac{{\left(\frac{A_1}{A_2}+1\right)}^2}{{\left(\frac{A_1}{A_2}-1\right)}^2}=\frac{{\left(3+1\right)}^2}{{\left(3-1\right)}^2}=4.$}