Q. No.Huygen's principle for secondary wavelets may be used to:
1.
explain Snell's law.
2.
find the velocity of light in vacuum.
3.
find a new position of a wavefront.
4.
both (1) & (3) are correct.
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| 1. | \(I_0\) | 2. | \(\dfrac{I_0}{2}\) |
| 3. | \(\dfrac{I_0}{4}\) | 4. | \(\dfrac{I_0}{8}\) |
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In Young's double-slit experiment, an electron beam is used to obtain an interference pattern. If the speed of electrons is increased:
| 1. | No interference pattern will be observed. |
| 2. | Distance between two consecutive fringes remains the same. |
| 3. | Distance between two consecutive fringes will decrease. |
| 4. | Distance between two consecutive fringes will increase. |
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The graph between resolving power and accelerating potential V for an electron microscope is (P is resolving power):
| 1. |  |
2. |  |
| 3. |  |
4. |  |
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Huygens' wave theory allows us to know the:
| 1. | wavelength of the wave. |
| 2. | velocity of the wave. |
| 3. | amplitude of the wave. |
| 4. | propagation of the wavefront. |
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Two coherent sources are \(0.3~\text{mm}\) apart. They are \(1~\text{m}\) away from the screen. The second dark fringe is at a distance of \(0.3~\text{cm}\) from the center. The distance of the fourth bright fringe from the centre is:
1. \(0.6~\text{cm}\)
2. \(0.8~\text{cm}\)
3. \(1.2~\text{cm}\)
4. \(0.12~\text{cm}\)
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If the ratio of amplitudes of two coherent sources producing an interference pattern is \(3:4\), the ratio of intensities at maxima and minima is:
1. \(3:4\)
2. \(9:16\)
3. \(49:1\)
4. \(25:7\)
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Young's double-slit experiment is performed in a liquid. The \(10\)th bright fringe in the liquid lies where the \(8\)th dark fringe lies in a vacuum. The refractive index of the liquid
is approximately:
1. \(1.81\)
2. \(1.67\)
3. \(1.54\)
4. \(1.33\)
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The resolving power of a microscope can be increased by using:
1. red light.
2. blue light.
3. oil between objective lens and object.
4. both (2) and (3).
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If the \(5\)th order maxima of wavelength \(4000~\mathring{A}\) in Young's double-slit experiment coincides with the \(n\)th order maxima of wavelength \(5000~\mathring{A},\) then \(n\) is equal to:
1. \(5\)
2. \(8\)
3. \(4\)
4. \(10\)
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