# Important Quantities Derivation from YDSE

Physics

## example

### Shift of fringes when a slab/lens in introduced in front of a slit

Example: transparent slabs of refractive index each, having thickness to are arranged one over another. A point object is seen through this combination with the perpendicular light. If the shift of object by the combination is then find the value of .

Solution:
As the refractive index of all the slabs are so those may be a considered as a single slab with the thickness summed up.
Summed up thickness of slabs
Given, for small angle of incidence, shift of object:
so,

So, or
As, can not be negative. so

## example

### Define and calculate fringe width and angular fringe width in YDSE

Fringe width is the distance between two successive bright fringes or two successive dark fringes. In the interference pattern, the fringe width is constant for all the fringes. Fringe width is independent of order of fringe. Fringe width is directly proportional to wavelength of the light used. It is given by:

Angular fringe width is given by:

Example: In Young's double slit experiment the two slits are illuminated by light of wavelength and the distance between the fringes obtained on the screen is . If the whole apparatus is immersed in water then find the angular fringe width. (refractive index of water is )

Solution:
For interference in YDSE:

For small , can be approximated to
So,
or,
Now as the set up immersed in water,  will become
So,
or,
or,

## example

### Find the ratio of maximum to minimum intensity

Example: In Young's double slit experiment, first slit has width four times the width of the second slit. Find the ratio of the maximum intensity to the minimum intensity in the interference fringe system.

Solution:

Let intensity from the smaller slit be . So, intensity from the bigger slit will be since it is times of smaller slit.
Now,

So,