Question

From the figure given below, the refractive index of medium B with respect to medium A (_Aμ_B) is:

पर्याय

• `(sin 45^circ)/(sin 30^circ)`

• `(sin 30^circ)/(sin 45^circ)`

• `(sin 45^circ)/(sin 60^circ)`

• `(sin 60^circ)/(sin 45^circ)`

Answer 1

`bb((sin 60^circ)/(sin 45^circ))`

Explanation:

In medium A, the light ray makes an angle of 30° with the interface. Therefore, the angle of incidence (i) in medium A is:

i = 90° − 30°

= 60°

In medium B, the light ray makes an angle of 45° with the interface. Therefore, the angle of refraction (r) in Medium B is:

r = 90° − 45°

= 45°

The refractive index of medium B with respect to medium A (_Aμ_B) is defined as the ratio of the sine of the angle in medium A to the sine of the angle in medium B:

_Aμ_B = `"sin (angle in medium A)"/"sin (angle in medium B)"`

= `(sin 60^circ)/(sin 45^circ)`