Question

In YDSE with monochromatic light, fringes are obtained on the screen placed at some distance from the slits. If screen is moved by $5\times {10}^{-2}$m towards the slits, fringe width changes by $3\times {10}^{-5}m$. If separation between the slits is ${10}^{-3}m$ and wavelength of light used is $1000y\mathring{A}$, then $y$ is (Assume distance between the two slits is very small than the distance between the slits and screen)

Answer 1

The fringe width obtained on the screen is given by
$\beta =\frac{D\lambda }{d}$,
where $D$ is the distance between the screen and the slits; $d$ is the distance between the slits and $\lambda$ is the wavelength.
Hence we can write change in the fringe width as
$\Delta \beta =\frac{D_2\lambda }{d}-\frac{D_1\lambda }{d}$
Given that the change in the fringe width is $3\times {10}^{-5}m$ when the screen is moved by $5\times {10}^{-2}m$.
i.e., $D_2-D_1=5\times {10}^{-2}m$ and $\Delta \beta =3\times {10}^{-5}m$
$⟹3\times {10}^{-5}=\frac{5\times {10}^{-2}}{{10}^{-3}}\lambda$
$⟹\lambda =6000A=1000\times 6A$
$⟹y=6$