Question

Tn a Young's double slit experiment, the intensity
at the central maximum is lo- The intensity at a
distance B/4 from the central maximum is (ß is
fringe width)
​

Answer 1

Answer:

1/2(Io)

Explanation:

Formulates used:-

B=(lambda)D/d =fringe width

I= (Io)[cos(ø/2)]^2

Where Io is intensity of central maxima

Refer the attachment for solution!

Hope this answer helped you

Answer 2

Answer:

⇒ Central max intensity = $\rm I_0$

$\implies \rm Fringe\; width \;(\beta) =\dfrac{\lambda* D}{d}$

The intensity at a  distance B/4 from the central maximum

$\implies \rm Y = \dfrac{\lambda * D}{4d}$

As,

$\implies \rm \Delta x = \dfrac{Y*d}{D}$

Therefore,

$\implies \rm \Delta x = \dfrac{\lambda*D}{4d} * \dfrac{d}{D}$

$\implies \rm \Delta x = \dfrac{\lambda}{4}$

Since,

$\rm \phi = \dfrac{2\pi}{\lambda} * \dfrac{\lambda}{4}$

$\blue{\boxed{\rm I = I_1 + I_2 + 2\sqrt{I_1*I_2*cos\phi} }}$

Here,

$\rm\implies \phi = 0$

$\implies I = I_0$

$\implies I_1+I_2=I_s$

$\rm I_0 = 4I_s\;--->(1)$

And,

$\rm cos\phi = \dfrac{\pi}{2}$

$\rm \implies I = 2I_s\;--->(2)$

From equation (1) and (2)

We get,

$\green{\boxed{\boxed{\rm {I = \dfrac{I_0}{2} }}}}$