Q54 / 75 · JEE Main 2025 (24 Jan Shift 2) · Physics

Young's double slit inteference apparatus is immersed in a liquid of refractive index 1.44. It has slit separation of 1.5 mm . The slits are illuminated by a parallel beam of light whose wavelength in air is 690 nm . The fringe-width on a screen placed behind the plane of slits at a distance of 0.72 m , will be :

  • Correct answer
    A
    0.23 mm
  • B
    0.33 mm
  • C
    0.63 mm
  • D
    0.46 mm

Solution

$\begin{aligned} & \Delta W=\frac{\lambda D}{d \cdot \mu}=\frac{690 \times 10^{-9} \times 0.72}{1.5 \times 10^{-3} \times 1.44} \\ & \Rightarrow \Delta W=2.3 \times 10^{-4} \mathrm{~m}=0.23 \mathrm{~mm}\end{aligned}$

Hence, the correct answer is option: 'A'.

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