Q. A beam of monochromatic blue light of wavelength $4200\ \mathring{\mathrm{A}}$ in air travels in water of refractive index $4/3$. Its wavelength in water will be
A) 4200 $\mathring{\mathrm{A}}$
B) 5800 $\mathring{\mathrm{A}}$
C) 4150 $\mathring{\mathrm{A}}$
D) 3150 $\mathring{\mathrm{A}}$
Explanation
(d) $\ _{air}\mu_{water} = \frac{speed\ of\ light\ in\ air}{speed\ of\ light\ in\ water} = \frac{c}{v}$
$\therefore\ _{air}\mu_{water} = \frac{v\lambda_{air}}{v\lambda_{water}}$
$\Rightarrow \lambda_{water} = \frac{\lambda_{air}}{\ _{air}\mu_{water}} = \frac{4200}{(4/3)} = \frac{3}{4} \times 4200 = 3150\ \mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.