Question

Young's double slit experiment is carried out using microwaves of wavelength $\lambda =3cm$. Distance between the slits is $d=5cm$ and the distance between the plane of slits and the screen is $D=100cm$. Then what is the number of maximas and their positions on the screen?

• A. $3,0$ and $\pm 75cm$
• B. $4,1$ and $\pm 60cm$
• C. $3,1$ and $\pm 55cm$
• D. $Noneofthese$

Answer 1

The correct option is A $3,0$ and $\pm 75cm$

The maximum path difference that can be produced = distance between the sources or $5cm$.

Thus, in this case we can have only three maximas, one central maxima and two on its either side for a path difference of $\lambda$ or $3cm$.

For maximum intensity at $P$,

$S_{2}P-S_{1}P=\lambda$

$\sqrt{(y+\frac{d}{2}{)}^2+D^2}-\sqrt{(y-\frac{d}{2}{)}^2+D^2}={\lambda }^2$

Solving the above equation by substituting $D=100cm$ and $d=5cm$ we get,

$y=\pm 75cm$

Thus the three maximas will be at $y=0$ and $y=\pm 75cm$

So the correct answer is option (a).