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Wave Optics question

2024 · 5 Apr · Shift 1 · Q89
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  5. /2024 · 5 Apr · Shift 1 · Q89

Wave Optics question

2024 · 5 Apr · Shift 1 · Q89

JEE MainPhysicsWave OpticsNumerical+4 / −1
In Young's double slit experiment, carried out with light of wavelength 5000 Ao5000~\mathop A\limits^o5000 Ao​, the distance between the slits is 0.3 mm0.3 \mathrm{~mm}0.3 mm and the screen is at 200 cm200 \mathrm{~cm}200 cm from the slits. The central maximum is at x=0 cmx=0 \mathrm{~cm}x=0 cm. The value of xxx for third maxima is ‾\underline{\hspace{2cm}}​mm\mathrm{mm}mm.
Numerical answer
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Correct answer: 10

  1. In Young’s double slit experiment, the position of the nnnth bright fringe (maxima) from the central maximum is xn=nβx_n=n\betaxn​=nβ where fringe width β=λDd\beta=\frac{\lambda D}{d}β=dλD​

  2. Convert all quantities into SI units:

  • Wavelength: λ=5000 A˚=5000×10−10 m=5×10−7 m\lambda=5000\,\mathring{A}=5000\times 10^{-10}\text{ m}=5\times 10^{-7}\text{ m}λ=5000A˚=5000×10−10 m=5×10−7 m
  • Slit separation: d=0.3 mm=0.3×10−3 m=3×10−4 md=0.3\text{ mm}=0.3\times 10^{-3}\text{ m}=3\times 10^{-4}\text{ m}d=0.3 mm=0.3×10−3 m=3×10−4 m
  • Screen distance: D=200 cm=2 mD=200\text{ cm}=2\text{ m}D=200 cm=2 m

  1. Calculate fringe width: β=λDd=(5×10−7)(2)3×10−4\beta=\frac{\lambda D}{d}=\frac{(5\times 10^{-7})(2)}{3\times 10^{-4}}β=dλD​=3×10−4(5×10−7)(2)​ β=10×10−73×10−4=10−63×10−4=13×10−2 m\beta=\frac{10\times 10^{-7}}{3\times 10^{-4}}=\frac{10^{-6}}{3\times 10^{-4}}=\frac{1}{3}\times 10^{-2}\text{ m}β=3×10−410×10−7​=3×10−410−6​=31​×10−2 m β=3.33×10−3 m=3.33 mm\beta=3.33\times 10^{-3}\text{ m}=3.33\text{ mm}β=3.33×10−3 m=3.33 mm

  2. For the third maximum: x3=3β=3×3.33 mm=9.99 mm≈10 mmx_3=3\beta=3\times 3.33\text{ mm}=9.99\text{ mm}\approx 10\text{ mm}x3​=3β=3×3.33 mm=9.99 mm≈10 mm

  3. Therefore, the position of the third maximum is x=10 mmx=10\text{ mm}x=10 mm

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