Wave Optics

JEE Main level, numerical answer, medium

82 of 191

Question 82 of 191

In a Young's double slit experiment, the slits are separated by 0.3mm0.3mm and the screen is 1.5m1.5m away from the plane of slits. Distance between fourth bright fringes on both sides of central bright fringe is 2.4cm.2.4cm. The frequency of light used is x×1014Hz.x\times {10}^{14}Hz.

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Answer: 5

The distance between nth{n}^{th} bright fringe on both side of central maxima is 2.4cm.2.4cm. The f=?f=?
Δy=Dλd×8\Delta y=\dfrac{D\lambda }{d}\times 8
λ=dΔy8D=0.3×10−3×2.4×10−28×1.5=0.12×10−52=0.06×10−5\lambda =\dfrac{d\Delta y}{8D}=\dfrac{0.3\times {10}^{-3}\times 2.4\times {10}^{-2}}{8\times 1.5}=\dfrac{0.12\times {10}^{-5}}{2}=0.06\times {10}^{-5}
λ=6×10−7m\lambda =6\times {10}^{-7}m
f=C6×10−7=0.5×10+15f=\dfrac{C}{6\times {10}^{-7}}=0.5\times {10}^{+15}
f=5×1014Hzf=5\times {10}^{14}Hz
So, x=5x=5

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