In Young's double-slit experiment, the slits are 0.4 mm apart and illuminated by photons of two wavelengths λ1 = 600 nm and λ2 = 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

Subtopic:  Young's Double Slit Experiment |
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A diffraction pattern is observed using a beam of red light. What will happen if the red light is replaced by the blue light?

1. No change takes place.
2. Diffraction bands become narrower.
3. Diffraction bands become broader.
4. Diffraction pattern disappears.

Subtopic:  Diffraction |
 81%
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The phenomenon of polarisation justify which nature of light?

1.  longitudinal 

2.  transverse

3.  both

4.  geometrical

 

Subtopic:  Polarization of Light |
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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

 

Subtopic:  Superposition Principle |
 75%
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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

Subtopic:  Resolving Power of Optical Devices (OLD NCERT) |
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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 Ao is used, the thickness (in μm) of mica is —

1.  90

2.  12

3.  14

4.  24

Subtopic:  Young's Double Slit Experiment |
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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

Subtopic:  Resolving Power of Optical Devices (OLD NCERT) |
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The distance of the moon from earth is 3.8×105 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

Subtopic:  Resolving Power of Optical Devices (OLD NCERT) |
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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. \(\dfrac{3}{4} I_{0}\)                                   

2. \(\dfrac{I_{0}}{8}\)

3. \(\dfrac{3}{8} I_{0}\)                                       

4. \(\dfrac{\sqrt{3}}{8} I_{0}\)

Subtopic:  Superposition Principle |
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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

Subtopic:  Resolving Power of Optical Devices (OLD NCERT) |
 61%
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