In a double-slit experiment, the two slits are \(1~\text{mm}\) apart and the screen is placed \(1~\text{m}\) away. Monochromatic light of wavelength \(500~\text{nm}\) is used. What will be the width of each slit for obtaining ten maxima of double-slit within the central maxima of a single-slit pattern?

1. \(0.2~\text{mm}\)
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

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.  $2\left({\mathrm{l}}_1+{\mathrm{l}}_2\right)$

2.  ${\left({\sqrt{\mathrm{I}}}_1+{\sqrt{\mathrm{l}}}_2\right)}^2$

3.  ${\left({\sqrt{\mathrm{I}}}_1-{\sqrt{\mathrm{l}}}_2\right)}^2$

4.  $2\left({\sqrt{\mathrm{I}}}_1-{\sqrt{\mathrm{l}}}_2\right)$

A major breakthrough in the studies of cells came with the development of an electron microscope. This is because:

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. \(0.05^{\circ}\)

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:

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. \(\dfrac{\pi}{8}~\text{radian}\)

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:

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. \(\dfrac{I_0}{4}\)

2. \(\dfrac{3}{4}I_0\)

3. \(\dfrac{I_0}{2}\)

4. \(I_0\)