Question

If the first minima in Young's double-slit experiment occurs directly in front of the slits (distance between slit and screen $D=12cm$ and distance between slits $d=5cm$), then the wavelength of the radiation used can be

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

Answer 1

Answer: (A)

$2cm$

Path travelled by light from slit $S_2$, $S_{2}P=\sqrt{D^2+d^2}=\sqrt{{12}^2+5^2}$

$S_{2}P=13cm$

Path travelled by light from slit $S_1$, $S_{1}P=D=12cm$

So the path difference,$\Delta x=S_{2}P-S_{1}P=1cm$

Now for the point P to be the first minima, $\Delta x=\frac{1}{2}\lambda$

So, $1=\frac{1}{2}\lambda$

$⟹\lambda =2cm$