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AIIMS 201926 May
Exam
AIIMS
Year
2019
Date
26 May
Subject
Physics
Chapter
Wave Optics

AIIMS 2019 · 26 May · Physics · Wave Optics

In a Young's double slit experiment, a light of wavelength 500 ~nm...

In a Young's double slit experiment, a light of wavelength 500 nm500 \mathrm{~nm} falls on it. Its slits separation is 2 mm2 \mathrm{~mm} and distance between plane of slits and screen is 2 m2 \mathrm{~m} then, find distance of a point on the screen from central maxima where intensity becomes 50%50 \% of central maxima.

Options

  1. A1000μm1000 \mu \mathrm{m}
  2. B500μm500 \mu \mathrm{m}
  3. C250μm250 \mu \mathrm{m}
  4. D125μm125 \mu \mathrm{m}

Correct answer

D. 125μm125 \mu \mathrm{m}

Step-by-step solution

 I=I0cos⁡2ϕ2=I02cos⁡ϕ2=12⇒ϕ2=π4,ϕ=π2Δx=λ4y=β4=λD4d=500×10−9×24×2×10−3=125×10−6my=125μm\begin{aligned} & \text{ } I=I_0 \cos ^2 \frac{\phi}{2}=\frac{I_0}{2} \\ & \cos \frac{\phi}{2}=\frac{1}{\sqrt{2}} \Rightarrow \frac{\phi}{2}=\frac{\pi}{4}, \phi=\frac{\pi}{2} \\ & \Delta x=\frac{\lambda}{4} \\ & y=\frac{\beta}{4}=\frac{\lambda D}{4 d}=\frac{500 \times 10^{-9} \times 2}{4 \times 2 \times 10^{-3}}=125 \times 10^{-6} \mathrm{ m} \\ & y=125 \mu \mathrm{m}\end{aligned}
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