Q64 / 75 · JEE Main 2026 (05 April Shift 1) · Physics
In Young's double slit experiment, the fringe width of the interference pattern produced on the screen is $2.4$ $\mu$m. If the experiment is carried out in another medium having refractive index $1.2$, the fringe width will be _____ $\mu$m.
- A$1.2$
- Correct answerB$2$
- C$2.4$
- D$2.88$
Solution
The fringe width in Young's double slit experiment is given by $\beta = \dfrac{\lambda D}{d}$.
When the experiment is performed in a medium of refractive index $\mu$, the wavelength of light becomes $\lambda' = \dfrac{\lambda}{\mu}$.
The new fringe width is $\beta' = \dfrac{\lambda' D}{d} = \dfrac{\lambda D}{\mu d} = \dfrac{\beta}{\mu}$.
Substituting the given values, $\beta = 2.4$ $\mu$m and $\mu = 1.2$:
$\beta' = \dfrac{2.4}{1.2} = 2$ $\mu$m.
Answer: $2$
When the experiment is performed in a medium of refractive index $\mu$, the wavelength of light becomes $\lambda' = \dfrac{\lambda}{\mu}$.
The new fringe width is $\beta' = \dfrac{\lambda' D}{d} = \dfrac{\lambda D}{\mu d} = \dfrac{\beta}{\mu}$.
Substituting the given values, $\beta = 2.4$ $\mu$m and $\mu = 1.2$:
$\beta' = \dfrac{2.4}{1.2} = 2$ $\mu$m.
Answer: $2$
Hence, the correct answer is option: 'B'.