Question

In Youngs double-slit experiment the distance between the centres of adjacent fringes is $0.10mm$. If the distance of the screen from the slits is doubled, the distance between the slits is halved and the wavelength of light is changed from $6.4\times 10^{-7}m$ to $4.0\times 10^{-7}m$, then the new distance between the fringes will be

• A. 0.10 mm
• B. 0.15 mm
• C. 0.20 mm
• D. 0.25 mm

Answer 1

Answer: (D)

$\beta =0.10mm$
$=0.10\times 10^{-3}m$
$\beta =\frac{\lambda D}{d}$
$\Rightarrow \frac{\beta }{\lambda }=\frac{D}{d}$
Again $\beta =\frac{4.0\times 10^7\times D}{\frac{d}{2}}$
$\beta =\frac{4.0\times 10^{-7}\times 4D}{d}$
$\beta =\frac{4.0\times 10^{-7}\times 4\times 0.10\times 10^{-3}}{6.4\times 10^{-7}}$
$\beta =0.25mm$