Wave Optics Test - 48

Given that,

Wavelength of light coming from the earth $$=\lambda $$

Wavelength of light coming from the distant star,$$\Delta \lambda =\lambda +\lambda \left( \dfrac{0.01}{100} \right)$$

Now,

  $$ \Delta \lambda =\lambda '-\lambda  $$

 $$ \Delta \lambda =\lambda +\lambda \left( \dfrac{0.01}{100} \right)-\lambda  $$

 $$ \Delta \lambda =\lambda \left( \dfrac{0.01}{100} \right) $$

  $$ \dfrac{\Delta \lambda }{\lambda }=\dfrac{v}{c} $$

 $$ v=3\times {{10}^{8}}\times {{10}^{-4}} $$

 $$ v=30\,km/s $$

Hence, the velocity of star is $$30\ km/s$$

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 $$6\,mW/{{m}^{2}}$$