JEE Main 2025 — Wave Optics Question with Solution

In Young's double-slit experiment, the width of the fringes, denoted as $\beta$, can be calculated using the formula:

$\beta =\frac{\lambda D}{d}$

In this formula:

$\lambda$ represents the wavelength of the light used.

$D$ is the distance from the slits to the screen.

$d$ is the distance between the slits.

According to this relationship, the fringe width $\beta$ is directly proportional to the wavelength $\lambda$. This means that as the wavelength increases, the fringe width also increases.

When comparing red and blue light:

Red light has a longer wavelength (${\lambda }_R$) compared to blue light (${\lambda }_b$).

Therefore, the fringe width ${\beta }_R$ for red light is greater than the fringe width ${\beta }_b$ for blue light, as ${\lambda }_R>{\lambda }_b$.

This shows that with red light, the fringes are actually farther apart compared to the fringes produced by blue light. Thus, the assertion that red light fringes are closer than those produced by blue light is false, while the reason about fringe width being proportional to the wavelength is true.