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. 0.5 m                        

2. 5 m

3. 5 mm                       

4. 5 cm

In a double-slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 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 mm

2. 0.1 mm

3. 0.5 mm

4. 0.02 mm

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 I_max/I_min is:
1. 9/4
2. 121/49
3. 49/121
4. 4/9

A beam of light of 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between first dark fringes on either side of the central bright fringe is
                      
1. 1.2cm

2. 1.2mm

3. 2.4cm

4. 2.4mm

Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are 

(1) 5I and I

(2) 5I and 3I

(3) 9I and I

(4) 9I and 3I

If the amplitude ratio of two sources producing interference is 3 : 5, the ratio of intensities at maxima and minima is 

(1) 25 : 16

(2) 5 : 3

(3) 16 : 1

(4) 25 : 9

For constructive interference to take place between two monochromatic light waves of wavelength λ, the path difference should be 

(1) $(2n-1)\frac{\lambda }{4}$

(2) $(2n-1)\frac{\lambda }{2}$

(3) $n\lambda$

(d) $(2n+1)\frac{\lambda }{2}$

Ray diverging from a point source forms a wavefront that is:

1. cylindrical.

2. spherical.

3. plane.

4. cubical.

In Young's double slit experiment, if the slit widths are in the ratio \(1:9,\) then the ratio of the intensity at minima to that at maxima will be:
1. \(1\)
2. \(1/9\)
3. \(1/4\)
4. \(1/3\)

The Young's experiment is performed with the lights of blue (λ = 4360 Å) and green colour (λ = 5460 Å), If the distance of the 4th fringe from the centre is x, then 

(1) x (Blue) = x (Green)

(2) x (Blue) > x (Green)

(3) x (Blue) < x (Green)

(4) $\frac{x\left(Blue\right)}{x\left(Green\right)}=\frac{5460}{4360}$