We have Young's double slit experiment. 
Distance between the slits, d = 2 mm = 2 × 10^-3 m
Distance between the slit and the screen, D = 140 cm = 1.40 m
Wavelength of the monochromatic source of light, λ = 600 nm = 600 × 10^-9m = 6 × 10^-7
m
Position of bright fringes is given by, 

               

Here, we are considering the third fringe.

∴ Distance of the third bright fringe is, 
           ${\mathrm{x}}_3 = 3\mathrm{\lambda }\frac{\mathrm{D}}{\mathrm{d}}$
$\Rightarrow$       ${\mathrm{x}}_3 = 3\times 6\times {10}^{-7}\times \frac{1.40}{2 \times {10}^{-3}}$
               $$\begin{gathered} = 12.6 \times {10}^{-4} \\ = 1.26 \times {10}^{-3} \mathrm{m} \\ = 1.26 \mathrm{mm} \end{gathered}$$  

If the wavelength of incident light is changed to 480 nm, then
 $\mathrm{\lambda } = 480 \mathrm{nm} = 480 \times {10}^{-9} = 4.8 \times {10}^{-7}\mathrm{m}$ 

Distance of the third bright fringe is

            $$\begin{gathered} {\mathrm{x}}_3 = 3\mathrm{\lambda }\frac{\mathrm{D}}{\mathrm{d}} \\ = 3 \times 4.8 \times {10}^{-7} \times \frac{1.40}{2 \times {10}^{-3}} \\ = 10.08 \times {10}^{-4} \\ = 1.008 \times {10}^{-3}\mathrm{m} \\ = 1.01 \times {10}^{-3}\mathrm{m} \\ = 1.01 \mathrm{mm}. \end{gathered}$$ 

$\therefore$ Shift in the position of third bright fringe when there is a change in wavelngth,
        $= 1.26 - 1.01 = 0.25 \mathrm{mm}.$

Here,
Distance between the slits, d = 0.3 mm = 0.3 × 10^-3 m
Fringe width, β = 1.5 mm = 1.5 × 10^-3 m
distance between screen and the slit, D = 7.5 cm = 0.75 m 

Wavelength λ is given by
                            $\boxed{\mathrm{\beta } = \frac{\mathrm{\lambda D}}{\mathrm{d}}}$ 

i.e.,                       $\mathrm{\lambda } = \frac{\mathrm{\beta d}}{\mathrm{D}}$
$\Rightarrow$           $$\begin{gathered} \mathrm{\lambda } = \frac{1.5 \times {10}^{-3}\times 0.3 \times {10}^{-3}}{0.75} \\ = 6000 \times {10}^{-10}\mathrm{m} = 6000 \mathrm{\AA } \end{gathered}$$ 


(a) Fringe width is given by, $\mathrm{\beta } = \frac{\mathrm{\lambda D}}{\mathrm{d}}$. 

when D is doubled, fringe width $\mathrm{\beta }$ is also doubled. 

i.e.,  $\mathrm{\beta } = 1.5 \times 2 = 3.0 \mathrm{mm}$ 

(b) When d is doubled, $\mathrm{\beta }$ is reduced to half. 

i.e.,      $\mathrm{\beta } = \frac{1.5}{2} = 0.75 \mathrm{mm}.$

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. 

Huygen's principle: (i) Every point on a given wavefront acts as a fresh source of secondary wavelets which travel in all directions with the speed of light.
(ii) The forward envelope of these secondary wavelets gives the new wavefront at any instant.
Laws of reflection by Huygen's principle: Let PQ be reflecting surface. Let a plane wavefront AB moving through the medium (air) towards the surface PQ meet at the point B. Let c be the velocity of light and t be the time of A to reach A' then AA' = ct.

By the Huygen's principle, secondary wavelets starts from B and cover a distance ct in time t and reaches at B'.
To obtain new wavefront, draw circles with point B as centre and ct (AA' = BB') as radius. Draw a tangent A'B' from the point A'.
Then A'B' represents the reflected wavelets which travels at right angle. Therefore, incident wavefront AB and reflected wavefront A'B' and normal lies in the same plane. In ∆ABA' and B'BA'
AA' = BB' = ct    [∵ AA' = BB' = BD = radii of same circle]
BA' = BA'    [common]
∠BAA' = ∆BB'A'    [each 90°]
∴ ∆ABA' ≅ ∠DBA'    [by R.H.S]
∠ABA' = ∠B'A'B    [C.P.C.T]
∴     incident angle i = reflected angle r
∠i = ∠r