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QJEEM18Pha2dd: JEE Main 2018 Physics - Class 12 JEEM18Pha2dd Solution: The angular width of the central maximum in a single slit…

The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 μ\mu m. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.)

Multiple Choice Options

  • A 100 μ\mu m
  • B 25 μ\mu m
  • C 50 μ\mu m
  • D 75 μ\mu m

Step-by-Step Verified Solution

JEE Main 2018 (Offline) Physics - Wave Optics Question 125 English Explanation

Given 2 θ\theta = 60o

⇒\Rightarrow θ\theta = 30o

We know,

aa sin θ\theta = n λ\lambda

for first minima n = 1,

∴\therefore At first minima

aa sin = λ\lambda

⇒\Rightarrow    \,\,\, 10 −- 6 ×\times sin 30o = λ\lambda

⇒\Rightarrow    \,\,\, λ\lambda = 10−62{{{{10}^{ - 6}}} \over 2} m

Now after making a new slit,

∴\therefore Fringe width, β\beta = λDd{{\lambda D} \over d}

given, D = 50 cm and β\beta = 1 cm.

∴\therefore 1 ×\times 10 −- 2 = 0.5×10−6×50×10−2d{{0.5 \times {{10}^{ - 6}} \times 50 \times {{10}^{ - 2}}} \over d}

⇒\Rightarrow d = 25 ×\times 10 −- 6 m = 25 μ\mu m.

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