In Young's double-slit experiment, the slits are 0.4 mm apart and illuminated by photons of two wavelengths ${\lambda }_1$ = 600 nm and ${\lambda }_2$ = 700nm. At what minimum distance from the common center, bright fringe from one interference pattern coincides with bright fringe from other? (Given that the distance of the screen from the plane of slits is 80cm.)

1.  7.2 mm

2.  6.0 mm

3.  8.4 mm

4.  9.8 mm

A diffraction pattern is observed using a beam of red light. What will happen if the red light is replaced by the blue light?

The phenomenon of polarisation justify which nature of light?

1.  longitudinal 

2.  transverse

3.  both

4.  geometrical

Two waves of intensity ratio 9:1 interfere to produce fringes in a young's double-slit experiment, the ratio of  intensity at maxima  to the intensity at minima is

1.  4:1

2.  9:1

3.  81:1

4.  9:4

If the wavelength of light used is halved and the numerical aperture of the compound microscope is doubled, then its resolving power will

1. Remain unchanged

2. Doubled

3. Halved

4. Quadrupled

Interference fringes are produced using white light in a double-slit arrangement. When a mica sheet of uniform thickness of the refractive index 1.6 (relative to air) is placed in the path of light from one of the slits, the central fringe moves by a distance.

This distance is equal to the width of 30 interference bands. If the light of wavelength 4800 $\stackrel{\mathrm{o}}{\mathrm{A}}$ is used, the thickness (in $\mu$m) of mica is —

1.  90

2.  12

3.  14

4.  24

An astronaut is looking down on earth's surface from a space shuttle at an altitude of

$400 km$. Assuming that the astronaut's pupil diameter is 5 mm and the wavelength of

visible light is 500 nm. The astronaut will be able to resolve linear object of the size of

about 

1. 0.5 m                   

2. 5 m

3. 50 m                     

4. 500 m

The distance of the moon from earth is $3.8\times {10}^5 km$. The eye is most sensitive to light of

wavelength 5500 Å. The minimum separation between two points on the moon that can

be resolved by a 500 cm telescope will be 

1. 51 m                         

2. 60 m

3. 70 m                         

4. All the above

A light source is located at $P_1$ as shown in the figure. All sides of the polygon are equal. The intensity of illumination at ${P_2}_{}$ is $I_0$ . What will be the intensity of illumination at $P_3$?

1. $\frac{3}{4}{I}_0$                                   

2. $\frac{I_0}{8}$

3. $\frac{3}{8}{I}_0$                                       

4. $\frac{\sqrt{3}}{8}{I}_0$

We wish to see inside an atom. Assuming the atom to have a diameter of 100 pm, this

means that one must be able to resolve a width of say 10 p.m. If an electron

microscope is used, the minimum electron energy required is about

1. 1.5 KeV                         

2. 15 KeV

3. 150 KeV                         

4. 1.5 KeV