Q. No.Two slits in Young’s experiment have widths in the ratio of \(1:25.\) The ratio of intensity at the maxima and minima in the interference pattern \(\dfrac{I_{max}}{I_{min}}\) is:
1.
\(\dfrac{9}{4}\)
2.
\(\dfrac{121}{49}\)
3.
\(\dfrac{49}{121}\)
4.
\(\dfrac{4}{9}\)
At the first minimum adjacent to the central maximum of a single slit diffraction pattern, the phase difference between the Huygen’s wavelet from the edge of the slit and the wavelet from the midpoint of the slit is:
1. \(\dfrac{\pi}{4}~\text{radian}\)
2. \(\dfrac{\pi}{2}~\text{radian}\)
3. \({\pi}~\text{radian}\)
4. \(\dfrac{\pi}{8}~\text{radian}\)
For a parallel beam of monochromatic light of wavelength \(\lambda\), diffraction is produced by a single slit whose width \(a\) is much greater than the wavelength of the light. If \(D\) is the distance of the screen from the slit, the width of the central maxima will be:
| 1. | \(\dfrac{2D\lambda}{a}\) | 2. | \(\dfrac{D\lambda}{a}\) |
| 3. | \(\dfrac{Da}{\lambda}\) | 4. | \(\dfrac{2Da}{\lambda}\) |
2. \(0.1~\text{mm}\)
3. \(0.5~\text{mm}\)
4. \(0.02~\text{mm}\)
In Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is \(\lambda\) is \(K\), (\(\lambda\) being the wavelength of light used). The intensity at a point where the path difference is \(\frac{\lambda}{4}\) will be:
1. \(K\)
2. \(\frac{K}{4}\)
3. \(\frac{K}{2}\)
4. zero
| 1. | The angular width of the central maximum of the diffraction pattern will increase. |
| 2. | The angular width of the central maximum will decrease. |
| 3. | The angular width of the central maximum will be unaffected. |
| 4. | A diffraction pattern is not observed on the screen in the case of electrons. |
Two periodic waves of intensities I1 and I2 pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is:
1.
2.
3.
4.
A major breakthrough in the studies of cells came with the development of an electron microscope. This is because:
| 1. | the resolution power of the electron microscope is much higher than that of the light microscope. |
| 2. | the resolving power of the electron microscope is 200-350 nm compared to 0.1-0.2 nm for the light microscope. |
| 3. | electron beam can pass through thick materials, whereas light microscopy requires thin sections. |
| 4. | the electron microscope is more powerful than the light microscope as it uses a beam of electrons that has a wavelength much longer than that of photons. |
In a double-slit experiment, when the light of wavelength \(400~\text{nm}\) was used, the angular width of the first minima formed on a screen placed \(1~\text{m}\) away, was found to be \(0.2^{\circ}.\) What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? \(\left(\mu_{\text{water}} = \dfrac{4}{3}\right)\)
1. \(0.1^{\circ}\)
2. \(0.266^{\circ}\)
3. \(0.15^{\circ}\)
4. \(0.05^{\circ}\)
The intensity at the maximum in Young's double-slit experiment is \(I_0\). The distance between the two slits is \(d= 5\lambda\), where \(\lambda \) is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance \(D = 10 d\)?
1. \(\frac{I_0}{4}\)
2. \(\frac{3}{4}I_0\)
3. \(\frac{I_0}{2}\)
4. \(I_0\)
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