As shown in the figure, in Young's double slit experiment, a thin plate of thickness $\mathrm{t}=10 \mu \mathrm{~m}$ and refractive index $\mu=1.2$ is inserted infront of slit $\mathrm{S}_{1}$. The experiment is conducted in air ($\mu=1$) and uses a monochromatic light of wavelength $\lambda=500 \mathrm{~nm}$. Due to the insertion of the plate, central maxima is shifted by a distance of $x\beta_{0}$. $\beta_{0}$ is the fringe-width before the insertion of the plate. The value of the x is \_\_\_\_.

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Wave Optics

As shown in the figure, in Young's double slit experiment, a thin plate of thickness

t = 10 μ m

and refractive index

μ = 1.2

is inserted infront of slit

S_{1}

. The experiment is conducted in air (

μ = 1

) and uses a monochromatic light of wavelength

λ = 500 nm

. Due to the insertion of the plate, central maxima is shifted by a distance of

x β_{0}

β_{0}

is the fringe-width before the insertion of the plate. The value of the x is ____.

Answer: 4

Inserting the plate adds an extra optical path

(μ - 1) t

in the beam from

S_{1}

, which shifts the central maximum by

\frac{( μ - 1 ) t}{λ}

fringe widths. Thus

x = \frac{( μ - 1 ) t}{λ} = \frac{( 1.2 - 1 ) ( 10 \times 1 0 ^{- 6} )}{500 \times 1 0 ^{- 9}} = \frac{2 \times 1 0 ^{- 6}}{5 \times 1 0 ^{- 7}} = 4

. The central maximum shifts by

4 β_{0}

, so

x = 4

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Integer answer