### Wave Optics Question 3

#### Question 3 - 29 January - Shift 1

In a Young’s double slit experiment, two slits are illuminated with a light of wavelength $800 nm$. The line joining $A_1 P$ is perpendicular to $A_1 A_2$ as shown in the figure. If the first minimum is detected at $P$, the value of slits separation ’ $a$ ’ will be :

The distance of screen from slits $D=5 cm$

(1) $0.4 mm$

(2) $0.5 mm$

(3) $0.2 mm$

(4) $0.1 mm$

## Show Answer

#### Answer: (3)

#### Solution:

#### Formula: YDSE

$A_2 P-A_1 P=\frac{\lambda}{2} \quad($ Condition of minima $)$

$\sqrt{D^{2}+a^{2}}-D=\frac{\lambda}{2}$

$D(1+\frac{a^{2}}{D^{2}})^{1 / 2}-D=\frac{\lambda}{2}$

$D(1+\frac{1}{2} \times \frac{a^{2}}{D^{2}})-D=\frac{\lambda}{2}$

$\frac{a^{2}}{2 D}=\frac{\lambda}{2} \Rightarrow a=\sqrt{\lambda . D}$

$=\sqrt{800 \times 10^{-6} \times 50}$

$a=0.2 mm$