In Young's double-slit experiment, monochromatic light of wavelength 600 m illuminates the pair of slits and produces an interference pattern in which two consecutive bright fringes are separated by 10 mm. Another source of monochromatic light produces the interference pattern in which the two consecutive bright fringes are separated by 8 mm. Find the wavelength of light from the second source.
What is the effect on the interference fringes if the monochromatic source is replaced by a source of white light?

Here, we are given young's double slit experiment.
Wavelength of monochromatic light, λ₁ = 600 nm = 600 × 10^-9 m
Fringe width, β₁ = 10 mm = 10 × 10^-3 m
Fringe width, β₂ = 8 mm = 8 × 10^-3 m 

Let d be the slit width and D the distance between slit and screen, then we have 

$$\begin{gathered} \mathrm{Fringe} \mathrm{width} \mathrm{due} \mathrm{to} \mathrm{first} \mathrm{source}, {\mathrm{\beta }}_1 = \frac{{\mathrm{\lambda }}_1\mathrm{D}}{\mathrm{d}} \mathrm{and}, \\ \mathrm{Fringe} \mathrm{width} \mathrm{due} \mathrm{to} \mathrm{second} \mathrm{source}, {\mathrm{\beta }}_2 = \frac{{\mathrm{\lambda }}_2\mathrm{D}}{\mathrm{d}} \end{gathered}$$ 

$\therefore$  $\frac{{\mathrm{\beta }}_1}{{\mathrm{\beta }}_2} = \raisebox{1ex}{${\displaystyle \frac{{\mathrm{\lambda }}_1\mathrm{D}}{\mathrm{d}}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \frac{{\mathrm{\lambda }}_2\mathrm{D}}{\mathrm{d}}}$}\right.$ 

$\Rightarrow$  $\frac{{\mathrm{\beta }}_1}{{\mathrm{\beta }}_2} = \frac{{\mathrm{\lambda }}_1}{{\mathrm{\lambda }}_2}$ 

$\Rightarrow$  $\frac{10 \times {10}^{-3}}{8 \times {10}^{-3}} = \frac{600 \times {10}^{-9}}{{\mathrm{\lambda }}_2}$ 

$\Rightarrow$  $\frac{10}{8} = \frac{600 \times {10}^{-9}}{{\mathrm{\lambda }}_2}$

i.e., ${\mathrm{\lambda }}_2 = \frac{8 \times 600 \times {10}^{-9}}{10} = 480 \times {10}^{-9} \mathrm{m}$
          $= 480 \mathrm{nm}.$ 

is the required wavelength of light from the second source.

If the monochromatic source is replaced by white light, then we will not be able to see the interference fringes because white light is not a coherent source of light. The condition for interference to take place is, the availabilty of coherent sources of light.