Question

In a young's double slit experiment, slits are illuminated by a monochromatic source of wavelength $6000\,\mathring{A}$ and fringes are obtained. If screen is moved by a distance of $5 \,cm$ towards slits, change in fringe width is $3 \times 10^{-5} m$. Then separation between the slits will be :

• A. $1 \,mm$
• B. $1.2 \,mm$
• C. $1.5\, mm$
• D. $1.63 \,mm$

Answer 1

Answer: (A)

Fringe widths $\beta$ and $\beta^{\prime}$ in a YDSE setup is given as
$ \beta=\frac{\lambda D}{d} $ and $ \beta^{\prime}=\frac{\lambda D^{\prime}}{d}$
$\Rightarrow \beta-\beta^{\prime}=\frac{\lambda\left(D-D^{\prime}\right)}{d}$
$\Rightarrow d=\frac{\lambda\left(D-D^{\prime}\right)}{\left(\beta-\beta^{\prime}\right)}$
$\Rightarrow d=\frac{6000 \times 10^{-10} \times 5 \times 10^{-2}}{3 \times 10^{-5}}$
$=10^{-3} \,m =1 \,mm$