Two slits are made 0.8 mm apart and the screen is placed 1.5 m away. What should be the width of each slit to obtain 8 maxima of the double slit pattern within the central maximum of the single slit pattern?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.2 mm
Each slit of width \( a \) produces a single-slit diffraction envelope, and the two slits together produce interference fringes of width \( \lambda D/d \) inside it. The width of the central maximum of the diffraction pattern is \( 2\lambda D/a \).
For 8 interference maxima to fit within it:
\( \displaystyle \frac{2\lambda D}{a} = 8 \times \frac{\lambda D}{d} \)
\( \lambda \) and \( D \) cancel:
\( \displaystyle a = \frac{2d}{8} = \frac{d}{4} = \frac{0.8\,\mathrm{mm}}{4} = 0.2\,\mathrm{mm} \)
The screen distance 1.5 m is not needed, since it cancels.