Question

In young’s double slit experiment the distance d between the slits ${S_{1}}_{1}$and $S_{2}$is 1 mm. What should the width of each slit be so to obtain 10 maxima of the double slit pattern within the central maximum of the single silt pattern?

• A. 0.9 mm
• B. 0.8 mm
• C. 0.2 mm
• D. 0.6 mm

Answer 1

Answer: (C)

0.2 mm

Answer 2

: The linear separation between n bright fringes in an interference pattern on the screen is $x_{n} = \frac{n\lambda D}{d}?$

As $x_{n} < < D,$ the angular separation between n bright fringes

$\theta_{n} = \frac{x_{n}}{D} = \frac{n\lambda}{d}$

For 10 bright fringes

$\theta_{10} = \frac{10\lambda}{d}$

The angular width of the central maximum in the diffraction pattern due to slit of width a is

$2\theta_{1} = \frac{2\lambda}{a}$

Now $\frac{10\lambda}{d} < \frac{2\lambda}{a}$or $a \leq \frac{d}{5} = \frac{1}{5}$mm $= 0.2mm$