Physics Q39 – JEE-MAIN 2026

In Young's double slit experiment, the fringe width of the interference pattern produced on the screen is

2.4 μ m

. If the experiment is carried out in another medium having refractive index

1.2

, the fringe width will be _______

μ m

Clue 1 of 3: Understanding Fringe Width

The fringe width in Young's double-slit experiment depends on the wavelength of light, slit separation, and screen distance.

🧠 Full Solution Path

Step 1: Relate Fringe Width to Wavelength and Refractive Index✦ Active

The fringe width in Young's double-slit experiment is given by $β = \frac{λ D}{d}$ , where $λ$ is the wavelength of light, $D$ is the distance between the slits and the screen, and $d$ is the distance between the slits. When the experiment is performed in a medium with refractive index $n$ , the wavelength of light changes from $λ_{a}$ (in air/vacuum) to $λ_{m} = \frac{λ_{a}}{n}$ . Consequently, the fringe width in the medium, $β_{m}$ , will be:

β_{m} = \frac{λ_{m} D}{d} = \frac{(λ_{a} / n) D}{d} = \frac{1}{n} (\frac{λ_{a} D}{d})

This simplifies to $β_{m} = \frac{β_{a}}{n}$ , where $β_{a}$ is the fringe width in air.

Step 2: Substitute Given Values○ Expand

Given the initial fringe width in air $β_{a} = 2.4 μ m$ and the refractive index of the new medium $n = 1.2$ . We substitute these values into the derived relationship:

β_{m} = \frac{2.4 μ m}{1.2}

Step 3: Calculate the New Fringe Width○ Expand

Performing the division, we find the new fringe width:

β_{m} = 2 μ m

The new fringe width is $2 μ m$ , which corresponds to option "2".

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Physics Question 39 – JEE-MAIN 2026

Clue 1 of 3: Understanding Fringe Width

🧠 Full Solution Path