In the Young's double slit experiment, the distance between the slits varies in time as d(t) = d? + a? sin(ωt); where d?, ω and a? are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :

At lower end
Tension, T? = 2g = 20 N (due to the 2 kg block)
Velocity, v? = √ (T? /μ) = √ (20/μ)
Wavelength, λ? = 6 cm

At upper end
Tension, T? = (2 kg + 6 kg)g = 8g = 80 N (due to the block and the rope)
Velocity, v? = √ (T? /μ) = √ (80/μ) = √4 * √ (20/μ) = 2v?

Since frequency (f) remains the same:
f = v? /λ? = v? /λ?
⇒ λ? = λ? * (v? /v? )
⇒ λ? = λ? * (2v? /v? ) = 2λ?
⇒ λ? = 2 * 6 cm = 12 cm

$I=I_{0}co{s}^{2}30°$

$=I_0{\left(\frac{\sqrt[]{3}}{2}\right)}^2=\frac{3}{4}{I}_0$ 

3d = 0.6mm

D = 80cm

= 800mm

Path difference is given by

BP – Andhra Pradesh = Dx

$\frac{}{}$ $\Rightarrow \frac{dy}{D}=\left(2n+1\right)\frac{\lambda }{2}$   [for Dark fringe at P]

n = 0, for first dark fringe

$\frac{dy}{D}=\frac{\lambda }{2}$

$\lambda =\frac{2d}{D}y$

$$ $=\frac{2d}{D}\times \frac{d}{2}[y=\frac{d}{2},Given$  first dark fringe is observed on the screen directly opposite to one of the slits]

$\lambda =\frac{2\times 0.6\times 0.6}{800\times 2}$

$\lambda =450mm$

$\frac{\text{Energy}}{\text{Volume}}=\frac{{\mathrm{M}\mathrm{L}}^2{\mathrm{T}}^{-2}}{{\mathrm{L}}^3}=\left({\mathrm{M}\mathrm{L}}^{-1}{\mathrm{T}}^{-2}\right)$

 The distance between two successive bright fringes is fringe width $\left(\beta \right)$ .

$\beta =\frac{\lambda D}{d}=\frac{589\times {10}^{-9}\times 1.5}{0.15\times {10}^{-3}}=5.9\mathrm{m}\mathrm{m}$

In primary rainbow, red colour is at top and violet is at bottom.

Intensity of secondary rainbow is less in comparison to primary rainbow.