Geometrical OpticsMedium
Question
The wavelength of light, while it is passing through water is 540 nm . The refractive index of water is $\frac{4}{3}$. The wavelength of the same light when it is passing through a transparent medium having refractive index of $\frac{3}{2}$ is $\_\_\_\_$ nm .
Options
A.380
B.840
C.480
D.540
Solution
$\left. \ \frac{\mu_{1}}{\mu_{2}} = \frac{v_{2}}{v_{1}} = \frac{\lambda_{2}}{\lambda_{1}} \right|\ v = f\lambda$
$${\frac{\mu_{1}}{\mu_{2}} = \frac{\lambda_{2}}{\lambda_{1}} }{\frac{\mu_{B}}{3/2} = \frac{\lambda}{540} }{\lambda = \left( \frac{4 \times 2}{3 \times 3} \times 540 \right) }{\lambda = 480\text{ }nm}$$
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