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Wave Optics question

2020 · 2 Sep · Shift 2 · Q59
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Wave Optics question

2020 · 2 Sep · Shift 2 · Q59

JEE MainPhysicsWave OpticsMCQ+4 / −1
In a Young’s double slit experiment, 16 fringes are observed in a certain segment of the screen when light of a wavelength 700 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen would be
  1. A
    28
  2. B
    24
  3. C
    30
  4. D
    18
View written solutionFree

Correct answer: A

  1. In Young’s double slit experiment, the fringe width is β=λDd\beta = \frac{\lambda D}{d}β=dλD​ where λ\lambdaλ is the wavelength.

  2. For the same experimental setup and the same screen segment, the number of fringes is inversely proportional to fringe width: N∝1β∝1λN \propto \frac{1}{\beta} \propto \frac{1}{\lambda}N∝β1​∝λ1​

  3. Therefore, N2N1=λ1λ2\frac{N_2}{N_1} = \frac{\lambda_1}{\lambda_2}N1​N2​​=λ2​λ1​​

  4. Given: N1=16,λ1=700 nm,λ2=400 nmN_1 = 16, \quad \lambda_1 = 700\,\text{nm}, \quad \lambda_2 = 400\,\text{nm}N1​=16,λ1​=700nm,λ2​=400nm So, N2=N1⋅λ1λ2=16⋅700400N_2 = N_1 \cdot \frac{\lambda_1}{\lambda_2} = 16 \cdot \frac{700}{400}N2​=N1​⋅λ2​λ1​​=16⋅400700​

  5. Calculate: N2=16⋅1.75=28N_2 = 16 \cdot 1.75 = 28N2​=16⋅1.75=28

  6. Thus, the number of fringes observed in the same segment is 28\boxed{28}28​

  7. Option check:

  • A: 282828 ✅
  • B: 242424 ❌
  • C: 303030 ❌
  • D: 181818 ❌

Hence, the correct option is A.

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