Question

in young's double slit experiment using light of wavelength 400 nm, interference fringes of width X are obtained the wavelength of light is increased to 600 NM and and the separation between the slits is halved if one wants the object of fringe width on the screen to be same in the two cases find the ratio of the distance between the screen and the plane of the interfering source in the two arrangements ​

Answer 1

$\boxed{\boxed{\mathtt{ Answer = 1.5 }}}$

Solution : Let ${D}_{1}$ be the distance between the screen and the sources, when light of wavelength 400nm is used.

Now, $\beta = \frac{D\:\lambda}{d}$

$\implies$ $\therefore$ $\frac{{D}_1 \times 400 \times 10^{- 9} }{d}$ = X ..... (i)

Let ${D}_{2}$ be the distance between the screen and the sources to obtain the same fringe width, when light of the wavelength 600nm is used. then,

$\implies$ $\frac{{D}_2 \times 600 \times 10^{- 9} }{d}$ = X ...... (ii)

from (i) and (ii) we have

$\implies$ $\frac{{D}_1 }{{D}_2}$ = $\frac{600\times 10^{ - 9}}{400 \times 10^{ - 9}}$ = 1.5