Q75 / 75 · JEE Main 2026 (28 January Shift 2) · Physics
A beam of light consisting of wavelengths 650 nm and 550 nm illuminates the Young's double slits with separation of 2 mm such that the interference fringes are formed on a screen, placed at a distance of 1.2 m from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is $\_\_\_\_$ $\times 10^{-5} \mathrm{~m}$.
Correct answer: 429
Solution
Given wavelengths are $\lambda_1 = 650$ nm and $\lambda_2 = 550$ nm.
Slit separation $d = 2$ mm $= 2 \times 10^{-3}$ m.
Distance to screen $D = 1.2$ m.
For the bright fringes of both wavelengths to coincide at a distance $y$ from the central maximum, we have:
$y = n_1 \frac{\lambda_1 D}{d} = n_2 \frac{\lambda_2 D}{d}$
where $n_1$ and $n_2$ are integers representing the order of the fringes.
This simplifies to $n_1 \lambda_1 = n_2 \lambda_2$.
$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{550}{650} = \frac{11}{13}$
The least values for coincidence are $n_1 = 11$ and $n_2 = 13$.
The least distance $y$ is given by:
$y = n_1 \frac{\lambda_1 D}{d} = 11 \times \frac{650 \times 10^{-9} \times 1.2}{2 \times 10^{-3}}$
$y = 11 \times 650 \times 0.6 \times 10^{-6}$ m
$y = 11 \times 390 \times 10^{-6}$ m
$y = 4290 \times 10^{-6}$ m
$y = 429 \times 10^{-5}$ m.
Slit separation $d = 2$ mm $= 2 \times 10^{-3}$ m.
Distance to screen $D = 1.2$ m.
For the bright fringes of both wavelengths to coincide at a distance $y$ from the central maximum, we have:
$y = n_1 \frac{\lambda_1 D}{d} = n_2 \frac{\lambda_2 D}{d}$
where $n_1$ and $n_2$ are integers representing the order of the fringes.
This simplifies to $n_1 \lambda_1 = n_2 \lambda_2$.
$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{550}{650} = \frac{11}{13}$
The least values for coincidence are $n_1 = 11$ and $n_2 = 13$.
The least distance $y$ is given by:
$y = n_1 \frac{\lambda_1 D}{d} = 11 \times \frac{650 \times 10^{-9} \times 1.2}{2 \times 10^{-3}}$
$y = 11 \times 650 \times 0.6 \times 10^{-6}$ m
$y = 11 \times 390 \times 10^{-6}$ m
$y = 4290 \times 10^{-6}$ m
$y = 429 \times 10^{-5}$ m.
Hence, the correct answer is 429.