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Standard XII

Physics

Phase Difference

In Young's do...

Question

In Young's double-slit experiment, the slit separation is $0.5 m m$ and the screen is $0.5 m$ away from the slit. For a monochromatic light of wavelength $500 n m$ , the distance of 3rd maxima from the 2nd minima on the other side of central maxima is

2.75 n m

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2.5 n m

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22.5 m m

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2.25 m m

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Solution

The correct option is B $2.25 m m$
Distance between third maxima and second minima on the other side,
$x = \frac{3 λ D}{d} + \frac{(2 \times 2 - 1) λ D}{2 d} = \frac{9 λ D}{2 d}$
$x = \frac{9 \times 500 \times 10^{- 9} \times 0.5}{2 \times 0.5 \times 10^{- 3}} = 2250 \times 10^{- 6} = 2.25 m m$

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In Young's double-slit experiment, the slit separation is 0.5mm and the screen is 0.5m away from the slit. For a monochromatic light of wavelength 500nm, the distance of 3rd maxima from the 2nd minima on the other side of central maxima is

The correct option is B 2.25mmDistance between third maxima and second minima on the other side,x=3λDd+(2×2−1)λD2d=9λD2dx=9×500×10−9×0.52×0.5×10−3=2250×10−6=2.25mm

In Young's double-slit experiment, the slit separation is $0.5 m m$ and the screen is $0.5 m$ away from the slit. For a monochromatic light of wavelength $500 n m$ , the distance of 3rd maxima from the 2nd minima on the other side of central maxima is

The correct option is B $2.25 m m$
Distance between third maxima and second minima on the other side,
$x = \frac{3 λ D}{d} + \frac{(2 \times 2 - 1) λ D}{2 d} = \frac{9 λ D}{2 d}$
$x = \frac{9 \times 500 \times 10^{- 9} \times 0.5}{2 \times 0.5 \times 10^{- 3}} = 2250 \times 10^{- 6} = 2.25 m m$