The diameter of the human eye lens is 2 mm. What will be the minimum distance between two points to resolve them, which are situated at a distance of 50 meters from the eye? The wavelength of light is 5000 Å:

1. 2.32 m

2. 4.28 mm

3. 1.52 cm

4. 12.48 cm

A telescope has an objective lens of 10 cm diameter and is situated at a distance of one kilometre from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is 5000 Å, is of the order of:

1. 5 m

2. 5 mm

3. 5 cm

4. 0.5 m

The angular resolution of a 10 cm diameter telescope for a wavelength of 5000 Å is of the order of:

1. 10^–4 rad

2. 10^–6 rad

3. 10⁶ rad

4. 10^–2 rad

A plane-polarized light with intensity \(I_0\) is incident on a polaroid with an electric field vector making an angle of \(60^{\circ}\) with the transmission axis of the polaroid. The intensity of the resulting light will be:

The first diffraction minima due to a single slit diffraction is at \(\theta = 30^{\circ}\) for a light of wavelength  \(5000~\mathring {A}.\)  The width of the slit is:
1. \(5\times 10^{-5}~\text{cm}\)
2. \(10\times 10^{-5}~\text{cm}\)
3. \(2.5\times 10^{-5}~\text{cm}\)
4. \(1.25\times 10^{-5}~\text{cm}\)

Light of wavelength \(6328~\mathring{A}\) is incident normally on a slit having a width of \(0.2\) mm. The width of the central maximum measured from minimum to minimum of diffraction pattern on a screen \(9\) m away will be nearly:
1. \(0.36^{\circ}\)
2. \(0.18^{\circ}\)
3. \(0.72^{\circ}\)
4. \(0.09^{\circ}\)

In Young's double-slit experiment, the slits are separated by \(0.28\) mm and the screen is placed \(1.4\) m away. The distance between the first dark fringe and the fourth bright fringe is obtained to be \(0.6\) cm. The wavelength of the light used in the experiment is:
1. \(3.4 \times 10^{-7}~\text{m}\)
2. \(4.1 \times 10^{-7}~\text{m}\)
3. \(3.4 \times 10^{-9}~\text{m}\)
4. \(4.1 \times 10^{-9}~\text{m}\)

Fringe width in a particular Young's double-slit experiment is measured to be \(\beta.\) What will be the fringe width if the wavelength of the light is doubled, the separation between the slits is halved and the separation between the screen and slits is tripled?
1. \(10\) times
2. \(11\) times
3. Same
4. \(12\) times

The relation between the fringe width for the red light and yellow light is: (all other things being the same.)
1. \(\beta_\text{red} < \beta_\text{yellow}\)
2. \(\beta_\text{red} > \beta_\text{yellow}\)
3. \(\beta_\text{red} = \beta_\text{yellow}\)
4. \(\beta_\text{red} =2 \beta_\text{yellow}\)

Two sources with intensity \(I_0\) and \(4I_0\) respectively interfere at a point in a medium. The maximum and the minimum possible intensity respectively would be:
1. \(2I_0, I_0\)
2. \(9I_0, 2I_0\)
3. \(4I_0, I_0\)
4. \(9I_0, I_0\)