Question

The wavelength of blue light in air is 4500${}^{\circ }$A.If the refractive index for blue light is 1.55 in glass, the frequency of the refracted light in glass and its wavelength are respectively ($c=3\times {10}^8$ m/s):

• A. $6.7\times {10}^{14}$ Hz, 2903A${}^{\circ }$
• B. $7.5\times {10}^{14}$ Hz, 3000A${}^{\circ }$
• C. $4.5\times {10}^{14}$ Hz, 3500A${}^{\circ }$
• D. $5.4\times {10}^{14}$ Hz, 7000A${}^{\circ }$

Answer 1

Answer: (A)

$6.7\times {10}^{14}$ Hz, 2903A${}^{\circ }$

If $i$ and $r$ is the angle of incidence and angle of reflection, then from Snell's law we have,
$\frac{sini}{sinr}=\frac{{\lambda }_1}{{\lambda }_2}=\frac{{\mu }_2}{{\mu }_1}$
where ${\lambda }_1$ and ${\lambda }_2$ are the wavelengths in media 1 and 2 respectively; and ${\mu }_1$ and ${\mu }_2$ are refractive indices of media 1 and 2 respectively.
For air, refractive index is $1$.
$⟹\frac{4500A}{{\lambda }_2}=\frac{1.55}{1}$
$⟹{\lambda }_2=2903A^0$

$f_2=\frac{v_2}{{\lambda }_2}=\frac{c}{\mu {\lambda }_2}=\frac{3\times {10}^8}{2903\times {10}^{-10}\times 1.55}$

Solving, $f_2=6.66\times {10}^{14}Hz$