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Wave Optics question

2016 · Q131
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Wave Optics question

2016 · Q131

NEETPhysicsWave OpticsMCQ+4 / −1
The intensity at the maximum in a Young's double slit experiment is III0. Distance between two slits is d = 5λ\lambdaλ, where λ\lambdaλ is the wavelength of light used in the expreriment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d ?
  1. A
    34I0{3 \over 4}{I_0}43​I0​
  2. B
    I02{{{I_0}} \over 2}2I0​​
  3. C
    I0
  4. D
    I04{{{I_0}} \over 4}4I0​​
View written solutionFree

Correct answer: B

NEET 2016 Phase 1 Physics - Wave Optics Question 32 English Explanation


Path difference S2P – S1P = $$\sqrt {{{\left( {50\lambda } \right)}^2} + {{\left( {5\lambda } \right)}^2}} $$ - 50$\lambda $ = 0.25$\lambda $

S2P – S1P = ${\lambda \over 4}$

Phase difference,

$\Delta $$\phi $ = $${{2\pi } \over \lambda } \times {\lambda \over 4}$$ = ${\pi \over 2}$

So, resultant intensity at the desired point 'P' is

I = I0cos2${\phi \over 2}$ = I0cos2${\pi \over 4}$ = ${{{I_0}} \over 2}$
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