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Wave Optics question

2020 · 3 Sep · Shift 1 · Q50
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Wave Optics question

2020 · 3 Sep · Shift 1 · Q50

JEE MainPhysicsWave OpticsMCQ+4 / −1
In a Young’s double slit experiment, light of 500 nm is used to produce an interference pattern. When the distance between the slits is 0.05 mm, the angular width (in degree) of the fringes formed on the distance screen is close to
  1. A
    0.17o
  2. B
    1.7o
  3. C
    0.57o
  4. D
    0.07o
View written solutionFree

Correct answer: C

  1. In Young’s double slit experiment, the angular fringe width is

Δθ=λd\Delta \theta = \frac{\lambda}{d}Δθ=dλ​

where:

  • λ=500 nm=500×10−9 m\lambda = 500\,\text{nm} = 500 \times 10^{-9}\,\text{m}λ=500nm=500×10−9m
  • d=0.05 mm=0.05×10−3 m=5×10−5 md = 0.05\,\text{mm} = 0.05 \times 10^{-3}\,\text{m} = 5 \times 10^{-5}\,\text{m}d=0.05mm=0.05×10−3m=5×10−5m
  1. Substitute the values:

Δθ=500×10−95×10−5\Delta \theta = \frac{500 \times 10^{-9}}{5 \times 10^{-5}}Δθ=5×10−5500×10−9​

Δθ=100×10−4=10−2 rad\Delta \theta = 100 \times 10^{-4} = 10^{-2}\,\text{rad}Δθ=100×10−4=10−2rad

So,

Δθ=0.01 rad\Delta \theta = 0.01\,\text{rad}Δθ=0.01rad

  1. Convert radians to degrees:

1 rad=180π∘1\,\text{rad} = \frac{180}{\pi}^\circ1rad=π180​∘

Hence,

0.01×180π≈0.01×57.3≈0.573∘0.01 \times \frac{180}{\pi} \approx 0.01 \times 57.3 \approx 0.573^\circ0.01×π180​≈0.01×57.3≈0.573∘

  1. Therefore, the angular width of fringes is approximately

0.57∘0.57^\circ0.57∘

  1. Checking options:
  • A: 0.17∘0.17^\circ0.17∘ ❌
  • B: 1.7∘1.7^\circ1.7∘ ❌
  • C: 0.57∘0.57^\circ0.57∘ ✅
  • D: 0.07∘0.07^\circ0.07∘ ❌

Therefore, the correct option is C.

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