JEE Physics 2026 Question with Solution

Given: $\lambda_1=650\,nm$, $\lambda_2=550\,nm$, $d=2\times10^{-3}\,m$, and $D=1.2\,m$.

Find: The least distance from the central maximum where bright fringes due to both wavelengths coincide.

In Young’s double slit experiment, the position of the $n$-th bright fringe is

$$y_n=\frac{n\lambda D}{d}$$

Bright fringes for two wavelengths coincide when

$$n_1\lambda_1=n_2\lambda_2$$

So,

$$650n_1=550n_2$$

which gives

$$13n_1=11n_2$$

The smallest integers satisfying this are

$$n_1=11,\quad n_2=13$$

Now the least common bright fringe position is

$$y=\frac{n_1\lambda_1 D}{d}$$

Substituting the values,

$$y=\frac{11\times650\times10^{-9}\times1.2}{2\times10^{-3}}$$

Thus,

$$y=4.29\times10^{-3}\,\text{m}$$

Therefore,

$$y=429\times10^{-5}\,\text{m}$$

So, the required numerical answer is 429.

Given: Two wavelengths $650\,nm$ and $550\,nm$ are used in YDSE with $d=2\,mm$ and $D=1.2\,m$.

Find: The first position where both bright fringes overlap.

For overlap of bright fringes, the path difference condition must be the same multiple of each wavelength. Hence,

$$n_1\lambda_1=n_2\lambda_2$$

Using the given wavelengths,

$$n_1\times650=n_2\times550$$

Dividing by $50$,

$$13n_1=11n_2$$

Since $13$ and $11$ are coprime, the least integer solution is

$$n_1=11,\quad n_2=13$$

Now use the fringe position formula for either wavelength:

$$y=\frac{n\lambda D}{d}$$

Taking $n_1=11$ and $\lambda_1=650\times10^{-9}\,\text{m}$,

$$y=\frac{11\times650\times10^{-9}\times1.2}{2\times10^{-3}}$$

Multiply the numerator:

$$11\times650\times1.2=8580$$

So,

$$y=\frac{8580\times10^{-9}}{2\times10^{-3}}$$ $$y=4290\times10^{-6}\,\text{m}=4.29\times10^{-3}\,\text{m}$$

Rewriting in the asked form,

$$y=429\times10^{-5}\,\text{m}$$

Therefore, the answer is 429.