Question

A ray of light travels from air to liquid by making an angle of incidence ${24}^o$ and angle of refraction is ${18}^o$. Find the R.I. of liquid. Determine the wavelength of light in air and in liquid the frequency of light is $5.4\times {10}^{15}$ Hz. c$=3\times {10}^8$ m/s.

• A. $2.316,5656$A; $422.13\stackrel{o}{A}$.
• B. $2.316,5556$A; $422.13\stackrel{o}{A}$.
• C. $1.316,5656$A; $422.13\stackrel{o}{A}$.
• D. $1.316,5556$A; $422.13\stackrel{o}{A}$.

Answer 1

The correct option is D $1.316,5556$A; $422.13\stackrel{o}{A}$.

Given angle of incidence ${\theta }_i$=${24}^0$ and angle of refraction ${\theta }_r$=${18}^0$ .

Also frequency of light in liquid =5.4$\times {10}^{15}$ and velocity of light in air =3 $\times {10}^8$.

From Snell's law,

$\frac{\sin {24}^0}{\sin {18}^0}=\frac{{\mu }_l}{{\mu }_a}$

Thus refractive index of liquid $=\frac{0.4067}{0.30901}=1.3161$

Also $\frac{{\mu }_l}{{\mu }_a}=\frac{v_a}{v_l}$----(A)

We know frequency f$=\frac{v}{\lambda }$

So equation A can be written as

$1.316=\frac{3\times {10}^8}{5.4\times {10}^{15}\times {\lambda }_L}{\lambda }_L=\frac{3\times {10}^8}{5.4\times {10}^{15}\times 1.316}=422.53\times {10}^{-10}$