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Wave Optics — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsWave Optics4 Marks
Question
When light from a monochromatic source is incident on a single narrow slit, it gets diffracted and a pattern of ahem ate bright and dark fringes is obtained on screen, called "Diffraction Pattern" of single slit. ln diffraction pattern of single slit, it is found that.
  1. Central bright fringe is of maximum intensity and the intensity of any secondary bright fringe decreases with increase in its order.
  2. Central bright fringe is twice as wide as any other secondary bright or dark fringe.
  1. A single slit of width 0.1mm is illuminated by a parallel beam oftight of wavelength $6000\mathring{\text{A}}$ and diffraction bands are observed on a screen 0.5m from the slit. The distance of the third dark band from the central bright band is:
  1. 3mm
  2. 1.5mm
  3. 9mm
  4. 4.5mm
  1. ln Fraunhofer diffraction pattern, slit width is 0.2mm and screen is at 2m away from the lens. If wavelength of tight used is $5000\mathring{\text{A}}$ then the distance between the first minimum on either side the central maximum is:
  1. $10^{-1}m$
  2. $10^{-2}m$
  3. $2 \times 10^{-2}m$
  4. $2 \times 10^{-1}m$
  1. Light of wavelength 600nm is incident normally on a slit of width 0.2mm. The angular width of central maxima in the diffraction pattern is (measured from minimum to minimum).
  1. $6 \times 10^{-3}$rad
  2. $4 \times 10^{-3}$rad
  3. $2.4 \times 10^{-3}$rad
  4. $4.5 \times 10^{-3}$rad
  1. A diffraction pattem is obtained by using a beam of red light. What will happen, if the red light is replaced by the blue light?
  1. Bands disappear
  2. Bands become broader and farther apart
  3. No change will take place
  4. Diffraction bands become narrower and crowded together.
  1. To observe diffraction, the size of the obstacle.
  1. Should be $\frac{\lambda}{2}$, where $\lambda$, is the wavelength.
  2. Should be of the order of wavelength.
  3. Has no relation to wavelength.
  4. Should be much larger than the wavelength.
✓

Answer

  1. (c) 9mm
Explanation:
Here, $\text{d} = 0.1 \text{mm},\lambda= 6000 \mathring{\text{A}}, \text{D}=0.5\text{m}$
For third dark band, $\text{d}\sin\theta=3\lambda\ ;\ \sin\theta=\frac{3\lambda}{\text{d}}=\frac{\text{y}}{\text{D}}$
$\text{y}=\frac{3\text{D}\lambda}{\text{d}}=\frac{3\times0.5\times6\times10^{-7}}{0.1\times10^{-3}}$
$=9\times10^{-3}\text{m}=9\text{mm}$
  1. (b) $10^{-2}m$
Explanation:
Given $d = 0.2mm = 0.2 \times 10^{-3}m, D = 2m$
$\lambda=5000\mathring{\text{A}}=5\times10^{-7}\text{m}$
The distance between the first minimum on other side of the central maximum
$\text{x}=\frac{2\lambda\text{D}}{\text{d}}=\frac{2\times5\times10^{-7}\times2}{0.2\times10^{-3}}$
$\Rightarrow\text{x}=10^{-2}\text{m}$
  1. (a) $6 \times 10^{-3}$rad
Explanation:
Here, $\lambda=600\text{nm}=6\times10^{-7}\text{m}$
$\text{a}=0.2\text{mm}=2\times10^{-4}\text{m},\theta=?$
Angular width of central maxima,
$\theta=\frac{2\lambda}{\text{a}}=\frac{2\times6\times10^{-7}}{2\times10^{-4}}=6\times10^{-3}\text{rad}$
  1. (d) Diffraction bands become narrower and crowded together.
Explanation:
When red light is replaced by blue light $(\lambda_\text{B}<\lambda_\text{R})$ the diffraction pattern bands becomes narrow and crowded together.
  1. (b) Should be of the order of wavelength.
Explanation:
To observe diffraction, the size of the obstacle should be of the order of wavelength.

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