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 (m = 1) will be equal to-

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

Answer 1

Answer: (B)

$\frac { \sin \theta } { \sin \phi ^ { \prime } }$

Answer 2

t = $\frac { \mathrm { BC } } { \mathrm { V } _ { \mathrm { a } } }$ = so, $\frac { \mathrm { BC } } { \mathrm { AD } }$ =

DACB BC = AC sinq

DACD AD = AC sinf¢

So == $\frac { \sin \theta } { \sin \phi ^ { \prime } }$

so $\frac { \mu _ { \mathrm { g } } } { \mu _ { \mathrm { a } } }$= $\frac { V _ { a } } { V _ { g } }$ = $\frac { \sin \theta } { \sin \phi ^ { \prime } }$