In a Young's double slit experiment, two slits are located 1.5 mm...

In a Young's double slit experiment, two slits are located 1.5 mm apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 nm. If the 20 maxima of the double slit pattern are contained within the centre maximum of the single slit diffraction pattern, then the width of each slit is

\mathrm{x} \times 10^{-3} \mathrm{~cm}

, where x -value is ________

Correct answer

Step-by-step solution

Width of 20 maxima of double slit = width of central maxima of single slit

\begin{aligned}<br/>& \frac{20 \lambda \mathrm{D}}{\mathrm{~d}}=\frac{2 \lambda \mathrm{D}}{\mathrm{a}} \\ & \frac{10}{\mathrm{~d}}=\frac{1}{\mathrm{a}} \\ & \mathrm{a}=\frac{\mathrm{d}}{10}=\frac{1.5 \times 10^{-1}}{10} \mathrm{~cm}=15 \times 10^{-3} \mathrm{~cm}<br/>\end{aligned}

Value of

x

is 15
Answer is 15

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