Question

In the adjoining diagram, a wavefront AB, moving in air is incident on a plane glass surface XY. Its position CD after refraction through a glass slab is shown also along with the normals drawn at A and D. The refractive index of glass with respect to air (μ=1 ) will be equal to

• A. $\frac{sin\theta }{sin{\theta }^{\prime }}$
• B. $\frac{sin\theta }{sin{\varphi }^{\prime }}$
• C. $\frac{sin{\varphi }^{\prime }}{sin\theta }$
• D. $\frac{AB}{CD}$

Answer 1

The correct option is B $\frac{sin\theta }{sin{\varphi }^{\prime }}$

in the case of refraction is CD is the refracted waves front and $V_1$ and $V_2$ are the speed of light in
the two media, then in the time the wavelets from B reaches C, the wavelets from A will reach D,
such that

$t=\frac{BC}{v_a}=\frac{AD}{v_g}\Rightarrow \frac{BC}{AD}=\frac{v_a}{v_g}...\left(i\right)$

But in $△ACB,BC=ACsin\theta ...\left(ii\right)$
While in $△ACD,AD=ACsin{\varphi }^{\prime }...\left(iii\right)$

From equations (i), (ii) and (iii) $\frac{v_a}{v_g}=\frac{sin\theta }{sin{\varphi }^{\prime }}$

Also $\mu \propto \frac{1}{v}\Rightarrow \frac{v_a}{v_g}=\frac{{\mu }_g}{{\mu }_a}=\frac{sin\theta }{sin{\varphi }^{\prime }}\Rightarrow {\mu }_g=\frac{sin\theta }{sin{\varphi }^{\prime }}$