Question

In the adjoining diagram,a wavelength 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 $(\mu =1)$ 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 if CD is the refracted wave 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 wavelet from A will reach D, such that,

$t=\frac{BC}{v_a}=\frac{AD}{v_g}⟹\frac{BC}{AD}=\frac{v_a}{v_g}$ ........(i)

But in $\Delta ACB,BC=ACsin\theta$ ..........(ii)

while in $\Delta ACD,AD=ACsin{\varphi }^{,}$ ............(iii)

From equation (i), (ii) and (iii)

$\frac{v_a}{v_g}=\frac{sin\theta }{sin{\varphi }^{\prime }}$

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