Q. In Young’s experiment, the third bright band for light of wavelength $6000\ \mathring{\mathrm{A}}$ coincides with the fourth bright band for another source of light in the same arrangement. Then the wavelength of second source is
A) $3600\ \mathring{\mathrm{A}}$
B) $4000\ \mathring{\mathrm{A}}$
C) $5000\ \mathring{\mathrm{A}}$
D) $4500\ \mathring{\mathrm{A}}$
Explanation
(d) $x = \frac{mD\lambda_{1}}{d} = \frac{(m + 1)D\lambda_{2}}{d}$
$\Rightarrow 3 \times 6000 = 4\lambda_{2}$
Or $\lambda_{2} = \frac{3\ \times \ 6000}{4} = 4500\ \mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.