In a Young's double-slit experiment…

In a Young's double-slit experiment with slit separation 0.50 mm and screen distance 1.0 m, the wavelength of the source in vacuum is 600 nm. The…

Numerical answer

In a Young's double-slit experiment with slit separation

0.50\ \mathrm{mm}

and screen distance

1.0\ \mathrm{m}

, the wavelength of the source in vacuum is

600\ \mathrm{nm}

. The whole apparatus is immersed in water (

\mu_w=4/3

). A thin transparent glass sheet of thickness

20\ \mu\mathrm{m}

and refractive index

1.5

is then placed in front of one of the slits (perpendicular to the beam). The number of fringes by which the central maximum shifts is ______. Give the answer correct to two decimal places.

Show answer and solution

Answer: 5.56

In water the wavelength is

\lambda_w=\lambda_0/\mu_w=450\ \mathrm{nm}

. The geometric path in water through the slab thickness

t

is replaced by a path in glass: the extra optical path is

(\mu_g-\mu_w)t

(it is relative to water, not to air). Converting to the equivalent path difference in water:

\Delta=(\mu_g/\mu_w-1)t=\left(\dfrac{1.5}{4/3}-1\right)(20\ \mu\mathrm{m})=0.125\times20=2.5\ \mu\mathrm{m}

. The number of fringes shifted is

N=\Delta/\lambda_w=\dfrac{2.5\times10^{-6}}{4.5\times10^{-7}}=5.56

(equivalently

(\mu_g-\mu_w)t/\lambda_0=\dfrac{(1/6)(20\times10^{-6})}{6\times10^{-7}}=5.56

). The shift of the central fringe is

N\beta

with

\beta=\lambda_wD/d=0.9\ \mathrm{mm}

, i.e. about

5\ \mathrm{mm}

. Using the air value

(\mu_g-1)t/\lambda_0=16.67

ignores the water.

Report a mistake

Next Ray and Wave Optics question →

Original question written in the JEE Advanced 2025 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake