Wave Optics

JEE Main level, single correct, medium

12 of 191

Question 12 of 191

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 :

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Answer: (A) 0.23 mm

ΔW=λDd⋅μ=690×10−9×0.721.5×10−3×1.44⇒ΔW=2.3×10−4 m=0.23 mm\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}

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