In a Young's double slit experiment, two slits are illuminated with a light of wavelength 800 nm . The line joining $\mathrm{A}_{1} \mathrm{P}$ is perpendicular to $\mathrm{A}_{1} \mathrm{~A}_{2}$ 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 $\mathrm{D}=5 \mathrm{~cm}$

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Wave Optics

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

(A)0.4 mm
(B)0.5 mm
(C)0.2 mm✓ Correct
(D)0.1 mm

Answer: (C)

Solution

Since

A_{1} P

is perpendicular to the slit line,

A_{1} P = D

while

A_{2} P = D^{2} + a^{2}

. The first minimum needs a path difference of

\frac{λ}{2}

, so

D^{2} + a^{2} - D = \frac{λ}{2}

. For

a ≪ D

the binomial expansion leaves

\frac{a ^{2}}{2 D} = \frac{λ}{2}

, hence

a = λ D = (800 \times 1 0^{- 9}) (0.05) = 2 \times 1 0^{- 4}

m. The slit separation is

0.2

mm.

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Single correct option

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