Explanation
It is given that,
$$ d=0.25\,cm $$
$$ D=100\,cm $$
$$ \lambda =6000\,\dot{A}=6\times {{10}^{-7}}\,m $$
For central maximum
$$ x=\dfrac{Dn\lambda }{d} $$
$$ x=\dfrac{100\times 6\times {{10}^{-7}}}{0.25} $$
$$ x=4\times {{10}^{-5}} $$
Intensity at this point is $${{I}_{0}}$$
At $$y=4\times {{10}^{-5}}\,m$$
Intensity will be $${{I}_{0}}$$
$$ \dfrac{{{I}_{\max }}}{{{I}_{\min }}}=\dfrac{{{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}}{{{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}^{2}}}=\dfrac{9}{1} $$
$$ \dfrac{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}=\dfrac{3}{1} $$
$$ \sqrt{\dfrac{{{I}_{1}}}{{{I}_{2}}}}=\dfrac{2}{1} $$
$$ \dfrac{{{I}_{1}}}{{{I}_{2}}}=\dfrac{4}{1} $$
So, the ratios of their intensities is $$4:1$$
To observe diffraction. the size of the obstacle
Hint :-The critical angle is that angle of incidence for which the angle of refraction in rarer medium is equal to 90.
Step 1: Note the given values and consideration
Let angle of incidence =$$\theta$$ ,
Angle of refraction =$$90^o$$,
Refractive index of liquid$$ =\mu$$,
Refractive index of air =1
Step 2: Calculate refractive index of liquid
Using Snell's law
$$ \mu \sin \theta =1 \times \sin90$$
$$ \mu \dfrac{3}{5}=1 $$
$$ \mu =\dfrac{5}{3} $$
Step 3: Calculate velocity of light in liquid
$$ \dfrac{c}{v}=\mu $$
$$ \dfrac{c}{v}=\dfrac{5}{3} $$
$$ v=\dfrac{3\times 3\times {{10}^{8}}}{5} $$
$$ v=1.8\times {{10}^{8}}m/s $$
Given that,
Intensity of central bright fringe $${{I}_{0}}=8mW/{{m}^{2}}$$
We know that,
$$ I={{I}_{0}}{{\cos }^{2}}\left( \frac{\phi }{2} \right) $$
$$ I=8{{\cos }^{2}}\left( \frac{\phi }{2} \right) $$
Now,
Phase difference = $$\dfrac{2\pi }{\lambda }$$ x path difference
$$ \phi =\dfrac{2\pi }{\lambda }\times \dfrac{\lambda }{6} $$
$$ \phi =\dfrac{\pi }{3} $$
Now, the intensity is
$$ I=8\times {{\cos }^{2}}\left( \dfrac{\pi }{6} \right) $$
$$ I=8\times \dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2} $$
$$ I=6\,mW/{{m}^{2}} $$
Hence, the intensity is $$6\,mW/{{m}^{2}}$$
Given,
As show in figure,
Plane wave front after reflection from concave mirror, turned into spherical wave front.
Hence, spherical wave front will appear.
For constructive
Path difference = $$n λ$$
$$ {{S}_{1}}F-{{S}_{2}}F=n\lambda $$
$$ 6m-4m=n\lambda $$
$$ 2\,m=n\lambda $$
$$ \lambda =\dfrac{n}{2}\,m $$
Where, $$n= 0, 1, 2….$$
So, $$\lambda =0,0.5\,m,1\,m,\dfrac{3}{2}\,m,2\,m....$$
Hence, the wave length is $$1\ m$$ in constructive interference
In the interference, two light waves are in the phase difference of at a point. If the yellow light is used, then the color of fringes at that point will be:-
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