Wave Optics Test - 41

Resolving power of telescope  RP$$=\dfrac{a}{1.22 \lambda}$$

(b) Resolving power $$\propto \displaystyle\frac{1}{\lambda}$$ and $$\propto \sqrt{v}$$
$$\therefore R.P\propto \sqrt{v}$$
$$\Rightarrow (R.P)_2=(R.P)_1\times \sqrt{\displaystyle\frac{80kV}{20kV}}$$
$$=2(R.P)_1$$.

By using the relation $$\cfrac { \Delta \lambda  }{ \lambda  } =\cfrac { a }{ c } ,a=c\times \cfrac { \Delta \lambda  }{ \lambda  } $$
$$=3\times { 10 }^{ 8 }\times \cfrac { 0.05 }{ 1000 } =1.5\times { 10 }^{ 5 }m/s$$
Also decrease in wavelength means the star is moving closer to the earth.

In Young's Double Slit Experiment (YDSE) be the position of central bright fringe on the screen in absence of any film, because the $$S_1O =S_2O$$.
When we introduce a film of thickness t and refractive index p, an additional path difference equal to $$(\mu t-t)=(\mu-1)t$$ is introduced. The optical path of upper bean becomes longer. For path difference on screen to be zero, path from lower slit $$S_2$$ should also be more. Thus, central bright fringe wire be located some what at O' as $$S_2O' > S_2O$$.
$$\therefore$$ The fringe pattern shifts upwards.
Now as a change in path difference of $$\lambda$$ corresponds to a change in position on the screen by $$\beta=\frac {D}{d} \lambda$$
Change in optical path difference $$\Delta y = (\mu - 1)t \times \frac {D}{d}$$
$$\Delta y= (1.5-1) ( 2\times 10^{-6}) \frac {D}{d}$$
$$\Delta y=\frac{1}{2} \times 2 \times 10^{-6} \frac {D}{d}=\frac {D}{d}\times 10^{-6} \, m$$
Fringe width $$=\frac {D}{d}\lambda =\frac {D}{d}\times 100 \times 10^{-9}m=\beta $$
Clearly, $$\frac{\Delta y}{\beta}=\frac{\frac {D}{d}\times 10^{-6}}{\frac {D}{d}\times 500 \times 10^{-9}}=2$$
$$= \frac{10^{-6}}{500 \times 10^{-9}}=\frac{10^{-6}\times 10^9}{500}=\frac{10^3}{500}$$
$$\therefore \Delta y= 2 \beta$$

When sources are coherent,  interference will take place and the intensity at mid - point of the screen will be $$ (a+a)^{2} = 4A^{2}$$. In case of incoherent sources, no interferencer will taake places and the intensity at mid - point will be $$A^{2}+A^{2}=2A$$. Hence the required ratio wil be $$2 : 1 $$

Resolving power of a microscope $$=\dfrac { 2\mu \sin { \theta  }  }{ 1.22\lambda  } $$
i.e., resolving power of a microscope is directly proportional to numerical aperture.

The condition of interference maxima is
$$d\sin\theta =n\lambda$$
$$\sin\theta =\displaystyle\frac{n\lambda}{d}$$
Given, $$d=2\lambda$$
$$\sin\theta =\displaystyle\frac{n\lambda}{2\lambda}=n/2$$
The magnitude of $$\sin\theta$$ lies between $$0$$ and $$1$$
When $$n=0, \sin \theta =0\Rightarrow \theta =0$$
When $$n=1, \sin\theta =1/2\Rightarrow \theta =30^o$$
When $$n=2, \sin\theta =1\Rightarrow \theta =90^o$$
Thus, there is central maximum $$(\theta =0^o)$$, on other side of it maxima lie at $$\theta =30^o$$ and $$\theta =90^o$$, so maximum number of possible interference maxima is $$5$$.