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Wave Optics — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsWave Optics4 Marks
Question
Interference is based on the superposition principle. According to this principle, at a particular point in the medium, the resultant displacement produced by a number of waves is the vector sum of the displacements produced by each of the waves.
If two sodium lamps illuminate two pinholes $S_1$ and $S_2,$ the intensities will add up and no interference fringes will be observed on the screen.
Here the source undergoes abrupt phase change in times of the order of $10^{-10}$ seconds.
  1. Two coherent sources of intensity $\text{10 }\frac{\text{W}}{\text{m}^2}$ and $\text{25 }\frac{\text{W}}{\text{m}^2}$ interfere to form fringes. Find the ratio of maximum intensity to minimum intensity.
  2. $\text{y}_1=\text{a}\sin\Big[\omega\text{t}+\frac{\pi}{3}\Big]$ and $\text{y}_2=\text{a}\sin\omega\text{t}$ is:
  3. $\text{a}$
  4. $\sqrt2\text{a}$
  5. $\text{2a}$
  6. $\sqrt3\text{a}$
  7. The resultant amplitude of a vibrating particle by the superposition of the two waves.
  8. Infinite
  9. Five
  10. Three
  11. Zero
  12. The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young's double $-$ slit experiment, is:
  13. $2D$
  14. $4D$
  15. $\frac{\text{D}}{2}$
  16. $\frac{\text{D}}{4}$
  17. ln a Young's double $-$ slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen $-$ to $-$ slit distance $D$ must be changed to:
  18. Soap bubble.
  19. Excessively thin film.
  20. A thick film.
  21. Wedge shaped film.
  22. Which of the following does not show interference?
  23. $15.54$
  24. $16.78$
  25. $19.72$
  26. $18.39$
✓

Answer

  1. $ 19.72$
Given $\text{I}_1=\text{10 }\frac{\text{W}}{\text{m}^2}$ and $\text{I}_2=\text{25 }\frac{\text{W}}{\text{m}^2}$
$\frac{\text{I}_1}{\text{I}_2}=\frac{\text{a}^2_1}{\text{a}^2_2}=\frac{10}{25}$
$\Rightarrow\frac{\text{a}_1}{\text{a}_2}=\frac{3.16}{5}$ or
$\text{a}_1=\frac{3.16}{5}\text{a}_2=0.6324\text{a}_2$
$\frac{\text{I}_\text{max}}{\text{I}_\text{min}}=\frac{(\text{a}_1+\text{a}_2)^2}{(\text{a}_1-\text{a}_2)^2}=\frac{[0.6324\text{a}_2+\text{a}_2]^2}{[0.6324\text{a}_2-\text{a}_2]^2}=19.724$
  1.  Excessively thin film.
In an excessively thin film, the thickness of the film is negligible.
Thus the path difference between the reflected rays becomes $\frac{\lambda}{2}$ which produces a minima.
  1.  $2D$
Since, $\beta=\frac{\lambda\text{D}}{\text{d}}$ for $d = 2d,$
$\beta' =\frac{\lambda\text{D}'}{2\text{d}}=\beta \ ($Gives$)$
$\therefore D_1 = 2D$
  1.  Five
The condition for possible interference maxima on the screen is, $\text{d}\sin\theta=\text{n}\lambda$
where $d$ is slit separation and $\lambda$ is the wavelength.
As $\text{d}=2\lambda \ ($given$)$ 
$\therefore2\lambda\sin\theta=\text{n}\lambda$ or $2\sin\theta=\text{n}$
For number of interference maxima to be maximum,
$\sin\theta=1\ \ \ \therefore\ \ \ \text{n}=2$
The interference maxima will be formed when
$\text{n}=0,\pm1,\pm2$
Hence the maximum number of possible maxima is $5.$
  1.  $\sqrt3\text{a}$
$\text{y}_1=\text{a}\sin\Big(\omega\text{t}\frac{\pi}{3}\Big)$ and $\text{y}_2=\text{a}\sin\omega\text{t}$
$\text{A}=\sqrt{\text{a}_1^2+\text{a}^2_2+2\text{a}_1\text{a}_2\cos\phi}$, where $\phi=\frac{\pi}{3}$
$=\sqrt{\text{a}^2+\text{a}^2+2\text{aa}\cos\frac{\pi}{3}}=\sqrt3\text{a}$

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