Question

In Youngs double-slit experiment the distance between the centres of adjacent fringes is $0.10 \,mm$. 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)

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