Physics Wave Optics

 $\frac{l_1}{I_2}=\frac{9}{4}$ $\frac{l_{max}}{l_{max}}=\frac{l_1+l_2+2\sqrt[]{l_1{l}_2}}{l_1+l_2-2\sqrt[]{l_1{l}_2}}$

$=\frac{\frac{9}{4}+1+2\sqrt[]{\frac{9}{4}}}{\frac{9}{4}+1-2\sqrt[]{\frac{9}{4}}}$

$=\frac{\frac{13}{4}+\frac{2*3}{2}}{\frac{13}{4}-2*\frac{3}{2}}=\frac{25}{1}$

.For maxima = y = (2n + 1) $\frac{\lambda }{2}\frac{D}{a}$ $\frac{}{}\frac{}{}$

For 1st maxima for l1 wavelength (n = 1)

${}_{}\frac{{}_{}}{}\frac{}{}$  $y_1=\frac{3{\lambda }_1}{2}\frac{D}{a}$ - (1)

First maxima for l2 wavelength

${}_{}$ $y_2=\frac{3}{2}{\lambda }_2\frac{D}{a}$  - (2)

$\Rightarrow y_2-y_1=\frac{3}{2}\frac{D}{a}\left[{\lambda }_2-{\lambda }_1\right]$

$=\frac{3}{2}*\frac{2*5*10^{-9}}{0.5*10^{-3}}$

$=\frac{3}{10^{-3}}*10^{-8}$

$\Delta y=3*10^{-5}m$

According to Young's double slit experiment, we can write

$\beta =\frac{\lambda D}{d}\Rightarrow \Delta \beta ={\beta }_2-{\beta }_1=\frac{\lambda }{d}\Delta D$

$\Rightarrow \lambda =\frac{d\Delta \beta }{\Delta D}=\frac{1*10^{-3}*3*10^{-5}}{5*10^{-2}}=60*10^{-8}m=600nm$

Given A₂ =   $\frac{A_1}{2}$

Bernoulli's b/w (1) & (2)

$\frac{P_1}{\rho g}+\frac{v_1^2}{2g}+z_1=\frac{{\rho }_2}{\rho g}+\frac{v_2^2}{2g}+z_2$

z₁ = z₂]

$\frac{P_1-P_2}{\rho g}=\frac{v_2^2-v_1^2}{2g}$

[Q = A₁V₁ = A₂V₂]

$\frac{4500}{750}=\frac{{\left(\frac{Q}{A_2}\right)}^2}{2}-{\left(\frac{Q}{A_1}\right)}^2$

$12=Q^2\left[\frac{1}{A_2^2}-\frac{1}{A_1^2}\right]$

$Q=2A_1=2*1.2*10^{-2}$  

= 2.4 * 10^-2 m³ /sec

= 24 * 10^-3 m³ /sec

= 24

Þ AC = d + t   $\left(\mu -1\right)\to$ path length                       

BC = d

path different = AC – BC =  $n\lambda$

$t\left(\mu -1\right)=\mu \lambda$

for contal maxima

$x\lambda \left(1.5-1\right)=n\lambda$

Þ   

0.5x = x

n = 0, 1,2- -

n = 0  Not possible

Now, n = 1

0.5x = 1

x = 2

${\beta }_1=\frac{{\lambda }_1{D}_1}{d_1}$

${\beta }_2=\frac{{\lambda }_2{D}_2}{d_2}$

$\frac{{\beta }_1}{{\beta }_2}=\frac{{\lambda }_1}{d_1}*\frac{d_2}{{\lambda }_2}$

$\frac{0.5}{{\beta }_2}=\frac{5000*d_1}{d_1*600}$

${\beta }_2=\frac{0.5*6}{10}$

2 = 0.3mm