Physics Q42 – JEE-MAIN 2026

The maximum intensity in a Young's double slit experiment is

I_{0}

. Distance between the slits (

d

) is

5 λ

, where

λ

is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at

D = 10 d

is _______.

Clue 1 of 3: Path Difference Calculation

Identify the geometry of the setup and determine the path difference between the waves from the two slits to the point of interest on the screen.

🧠 Full Solution Path

Step 1: Determine the path difference at the point of interest✦ Active

Let the two slits be $S_{1}$ and $S_{2}$ , separated by distance $d$ . Let $S_{2}$ be at $y = d / 2$ and $S_{1}$ at $y = - d / 2$ . The point $P$ on the screen is exactly opposite one of the slits, so we consider $P$ at $y = d / 2$ . The screen is at a distance $D$ from the slits.

The distance from $S_{2}$ to $P$ is $L_{2} = D$ (straight line path).

The distance from $S_{1}$ to $P$ is $L_{1} = \sqrt{D^{2} + (d / 2 - (- d / 2))^{2}} = \sqrt{D^{2} + d^{2}}$ .

The path difference is $Δ x = L_{1} - L_{2} = \sqrt{D^{2} + d^{2}} - D$ .

Step 2: Apply binomial approximation and substitute values○ Expand

Given $D = 10 d$ , which means $D ≫ d$ . We can use the binomial approximation $\sqrt{A^{2} + B^{2}} \approx A (1 + \frac{1}{2} \frac{B^{2}}{A^{2}})$ for $B ≪ A$ . Here, $A = D$ and $B = d$ .

Δ x \approx D (1 + \frac{1}{2} \frac{d^{2}}{D^{2}}) - D = D + \frac{d^{2}}{2 D} - D = \frac{d^{2}}{2 D}

Substitute the given values: $d = 5 λ$ and $D = 10 d = 10 (5 λ) = 50 λ$ .

Δ x = \frac{(5 λ)^{2}}{2 (50 λ)} = \frac{25 λ^{2}}{100 λ} = \frac{λ}{4}

💡 Teacher's Secret Hint

Ensure the approximation is valid by checking if $d / D$ is small. Here $d / D = 1 / 10$ , so $(d / D)^{2} = 1 / 100$ , which is small enough.

Step 3: Calculate the phase difference and intensity○ Expand

The phase difference $ϕ$ is related to the path difference $Δ x$ by $ϕ = \frac{2 π}{λ} Δ x$ .

ϕ = \frac{2 π}{λ} (\frac{λ}{4}) = \frac{π}{2}

The intensity $I$ at a point in Young's double-slit experiment is given by $I = I_{0} \cos^{2} (\frac{ϕ}{2})$ , where $I_{0}$ is the maximum intensity.

I = I_{0} \cos^{2} (\frac{π / 2}{2}) = I_{0} \cos^{2} (\frac{π}{4})

Since $\cos (\frac{π}{4}) = \frac{1}{\sqrt{2}}$ :

I = I_{0} {(\frac{1}{\sqrt{2}})}^{2} = I_{0} (\frac{1}{2}) = \frac{I_{0}}{2}

💡 Teacher's Secret Hint

Remember the relationship between path difference, phase difference, and intensity in interference patterns.

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Physics Question 42 – JEE-MAIN 2026

Clue 1 of 3: Path Difference Calculation

🧠 Full Solution Path