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QJEEM23Ph2542: JEE Main 2023 Physics - Class 12 JEEM23Ph2542 Solution: In a Young's double slit experiment, two slits are illuminated…

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:

JEE Main 2023 (Online) 29th January Morning Shift Physics - Wave Optics Question 42 English

The distance of screen from slits D = 5 cm

Multiple Choice Options

A 0.2 mm
B 0.5 mm
C 0.4 mm
D 0.1 mm

Step-by-Step Verified Solution

A_{2} P-A_{1} P=\frac{\lambda}{2} \quad

(Condition of minima)

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

\begin{aligned} D {(1 + \frac{a^{2}}{D^{2}})}^{1 / 2} - D = \frac{λ}{2} \\ D (1 + \frac{1}{2} \times \frac{a^{2}}{D^{2}}) - D = \frac{λ}{2} \\ \frac{a^{2}}{2 D} = \frac{λ}{2} \Rightarrow a = \sqrt{λ \cdot D} \\ = \sqrt{800 \times 10^{- 6} \times 50} \end{aligned}

\mathrm{a}=0.2 \mathrm{~mm}

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