Question

Derive the formula for the fringe width in young double slit experiment

Answer 1

To derive:

Fringe Width in YDSE:

Solution:

In YDSE , we can say :

Position of $n^{th}$ bright fringe:

$\rm d_{1} = \dfrac{n \lambda D}{d} \: \: \: \: . \: . \: . \: .(1)$

Now, position of $(n+1)^{th}$ bright fringe:

$\rm d_{2} = \dfrac{(n + 1) \lambda D}{d} \: \: \: \: . \: . \: . \: .(2)$

So, fringe width will be distance between adjacent bright fringes:

$\rm \beta = d_{2} - d_{1}$

$\rm \implies \beta = \dfrac{(n + 1) \lambda D}{d} - \dfrac{n \lambda D}{d}$

$\rm \implies \beta = \dfrac{\lambda D}{d}$

So, final answer is :

$\boxed{ \bf \beta = \dfrac{\lambda D}{d}}$

Answer 2

To Derive:

The formula for the fringe width in young doubt slit experiment (YDSE).

Solution:

We know that,

Position of $\sf n^{th}$ bright fringe is:

$\qquad \qquad \sf d_1 = \dfrac{n \lambda D}{d} \ . \ . \ . (i)$

Also,

Position of $\sf (n + 1)^{th}$ bridge fringe is:

$\qquad \qquad \sf d_2 = \dfrac{(n+1) \lambda D}{d} . \ . \ . (ii)$

Therefore,

Fridge width = Distance between adjacent bridge fringes:

$\qquad \quad \longrightarrow \sf \beta = d_2 - d_1$

$\qquad \quad \longrightarrow \sf \beta = \dfrac{(n-1) \lambda D}{d} - \dfrac{n \lambda D}{d}$

$\qquad \quad \longrightarrow \sf \beta = \dfrac{\lambda D}{d}$

Hence:

$\qquad \qquad \dag {\red{\underline{\boxed{\sf \beta = \dfrac{\lambda D}{d}}}}}$