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Wave Optics — Physics STD 12 Science — Question

Gujarat 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.
  1. $15.54$
  2. $16.78$
  3. $19.72$
  4. $18.39$
  1. Which of the following does not show interference?
  1. Soap bubble.
  2. Excessively thin film.
  3. A thick film.
  4. Wedge shaped film.
  1. 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:
  1. 2D
  2. 4D
  3. $\frac{\text{D}}{2}$
  4. $\frac{\text{D}}{4}$
  1. The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young's double-slit experiment, is:
  1. Infinite
  2. Five
  3. Three
  4. Zero
  1. The resultant amplitude of a vibrating particle by the superposition of the two waves.
$\text{y}_1=\text{a}\sin\Big[\omega\text{t}+\frac{\pi}{3}\Big]$ and $\text{y}_2=\text{a}\sin\omega\text{t}$ is:
  1. $\text{a}$
  2. $\sqrt2\text{a}$
  3. $\text{2a}$
  4. $\sqrt3\text{a}$
✓

Answer

  1. (c) 19.72
Explanation:
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. (b) Excessively thin film.
Explanation:
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. (a) 2D
Explanation:
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. (b) Five
Explanation:
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. (d) $\sqrt3\text{a}$
Explanation:
$\text{y}_1=\text{a}\sin\Big(\omega\text{t}\frac{\pi}{3}\Big)$ 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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