Question

A ray of light is incident at the glass–water interface at an angle i, it emerges finally parallel to the surface of water, then the value of ${\mu }_g$ would be

• A. $\left(\frac{4}{3}\right)sini$
• B. $\frac{1}{sini}$
• C. $\frac{4}{3}$
• D. $1$

Answer 1

The correct option is B $\frac{1}{sini}$

For glass water interface ${}_g{\mu }_{\omega }=\frac{sini}{sinr}.....\left(i\right)$ and For water-air interface ${}_{\omega }{\mu }_a=\frac{sinr}{sin90}......\left(ii\right)$
${\therefore }_g{\mu }_{\omega }{x}_{\omega }{\mu }_a=sini\Rightarrow {\mu }_g=\frac{1}{sini}$