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**7. (a) **What changes will occur to the interference fringe in Young's double slit experiment when

(i) distance between the slits is reduced and (ii) the whole system is immersed into water?

1)

Using the formula Y= λD/d

where D = distance from screen

d = distance between slits.

So fringe width will be increases

2) On immersing the apparatus in a liquid(say water), the wavelength of light decreases.

That is :-

λ'=$\frac{\lambda}{\mu}$ ..............(1)

As fringe width is calculated by:-

β=$\frac{\lambda D}{2d}$ ........(1)

Put value of λ=λ'μ in eq.(1):-

Now:- β'=$\frac{\lambda}{\mu}$*$\frac{D}{2d}$

=β/μ'

That is fringe width decreases and hence becomes $\frac{1}{\mu}$ times of its value in air.

Regards

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