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 mathrmx times 10^-3text cm$\mathrm{x} \times 10^{-3}\text{ cm}$, where x-value is ________.

### Related Formula Width of central maximum in single-slit diffraction: Delta y_textdiff = frac2lambda Da$$\Delta y_{\text{diff}} = \frac{2\lambda D}{a}$$ Fringe width in double-slit interference: beta = fraclambda Dd$$\beta = \frac{\lambda D}{d}$$ ### Core Logic Given condition: 20$20$ interference fringes fit inside the central diffraction envelope: 20 times beta = Delta y_textdiff$$20 \times \beta = \Delta y_{\text{diff}}$$ 20 times fraclambda Dd = frac2lambda Da$$20 \times \frac{\lambda D}{d} = \frac{2\lambda D}{a}$$ Cancel common parameters: frac10d = frac1a implies a = fracd10$$\frac{10}{d} = \frac{1}{a} \implies a = \frac{d}{10}$$ ### Step 1: Substitute Given Parameters Slit separation d = 1.5text mm = 0.15text cm$d = 1.5\text{ mm} = 0.15\text{ cm}$. a = frac0.15text cm10 = 0.015text cm = 15 times 10^-3text cm$$a = \frac{0.15\text{ cm}}{10} = 0.015\text{ cm} = 15 \times 10^{-3}\text{ cm}$$ Comparing with x times 10^-3text cm$x \times 10^{-3}\text{ cm}$, the value of x$x$ is **15**. ### Pattern Recognition Envelope matching conditions rely strictly on the geometric ratio of slit separation (d$d$) to individual slit width (a$a$). Wavelength (lambda$\lambda$) and screen distance (D$D$) cancel out completely. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics