Question

If the 8th bright band due to light of wavelength $\lambda_{1}$ coincides with 9 th bright band from light of wavelength $\lambda_{2}$ in Young's double slit experiment, then the possible wavelengths of visible light are

• A. 400 nm and 450 nm
• B. 425 nm and 400 nm
• C. 400 nm and 425 nm
• D. 450 nm and 400 nm

Answer 1

Answer: (D)

Position of $n$th bright fringe from central maxima is $\frac{n \lambda D}{d}$.
$\therefore \frac{8 \lambda_{1} D}{d} =\frac{9 \lambda_{2} D}{d}$
$\Rightarrow \frac{\lambda_{1}}{\lambda_{2}} =\frac{9}{8}$
Hence, the possible wavelengths of visible light is of the ratio of $9: 8$.