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The maximum number of possible interfere...

The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is

A

infinite

B

five

C

three

D

zero

Text Solution

Verified by Experts

The correct Answer is:
B

`dsintheta=nlamdaimpliesn=(dsintheta)/(lamda)=(2lamdasintheta)/(lamda)=2sintheta`
`impliesn_(max)=2`
`implies` Maximum number of possible Interference maxima `=(2n_(max)+1)=5`
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ALLEN -WAVE OPTICS-Exercise 5 A (Previous Years Questions)
  1. To demonstrate the phenomenon of interference, we require two sources ...

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  2. The maximum number of possible interference maxima for slit-separation...

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  3. A Young's double slit experiment uses a monochromatic source. The shap...

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  4. In a Young's double slit experiment the intensity at a point where tha...

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  5. A micture of light, consisting of wavelength 600nm and an unknown wave...

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  6. An intially parallel cyclindrical beam travels in a medium of refracti...

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  7. An intially parallel cyclindrical beam travels in a medium of refracti...

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  8. An intially parallel cyclindrical beam travels in a medium of refracti...

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  9. At two point P and Q on screen in Young's double slit experiment, wave...

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  10. In Young's double slit experiment, the two slits act as coherent sourc...

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  11. Statement-1: When light reflects from the air-glass plate interface, t...

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  12. In Young's double slit experiment, one of the slit is wider than other...

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  13. Two coherent point sources S1 and S2 are separated by a small distance...

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  14. Two beams A and B, of plane polarized light with mutually perpendicula...

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  15. On a hot summer night, the refractive index of air is smallest near th...

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  16. In a Young's double slit experiment with light of wavelength lamda the...

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  17. In Young's double slit experiment, the distance between slits and the ...

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