Question

The intensity of the central maximum in Youngs double-slit experiment is $4I$. The intensity at the first minimum is zero and the distance between two consecutive maxima is $w$. The distance from the central maximum to the position where the intensity falls to $I$ is

• A. $\frac{2}{3}\omega$
• B. $\frac{1}{4}\omega$
• C. $\frac{1}{2}\omega$
• D. $\frac{1}{3}\omega$

Answer 1

Answer: (D)

$\frac{1}{3}\omega$

Let the central maximum intensity be I1 i.e I1 = 4I

And the second intensity be I2 i.e I2 = I

According to the formula x = ($\frac{D}{d}$)$\lambda$ = $\omega$

Let $\beta$ = $\omega$ Which is the bandwidth

Therefore $\varphi$ = $\frac{2\pi }{\lambda }$.$\delta$x

$\frac{2\pi }{3}$ = $\frac{\delta x}{\lambda }$

Intensity I = 4I Cos${}^2$($\frac{\varphi }{2}$)

Putting square root on both sides we get;

$\frac{1}{2}$ = Cos($\frac{1}{2}$)

$\frac{\pi }{3}$ = $\frac{\varphi }{2}$

Therefore $\varphi$ = $\frac{2\pi }{3}$

$\frac{\lambda }{3}$ = $\delta$x

$\frac{\lambda }{3}$ =$\frac{xd}{D}$

Therefore $\frac{D\lambda }{d}$ = $\frac{x}{3}$

Hence x =$\frac{1}{3}$$\omega$