The diagram above shows an incident ray A, inclined at angle of 50^o to the interface CB. The ray OB is found to lie along the surface. What is the refractive index of the medium X with respect to air?

seaneverdries

10 years ago

n= sin i / sin r

draw the angles full indicate angle of incident and angle of refraction

:::

angle of refraction r = 90-50 = 40

and incident angle is 90

according to snell's law

n= sin90 / sin40

St3ssy

11 months ago

A normal LM is drawn through O as shown in the diagram above
Angle of incidence i = angle AOM = 90°-50°= 40°
Angle of refraction r from air = angle BOM = 90°
If you were asked to find the refractive index of air, your answer would be:
n = sin of angle of incidence in medium X /
sin of angle of refraction in air
= sin40°\sin90°
However, you are asked to find the refractive index of medium X with respect to air. So, based on the principle of reversibility of light, the refracted ray becomes the incident ray, and the incident ray becomes the refracted ray.

Retractive index of medium X = sin of angle of incidence in air/sin of angle of refraction in X
= sin90°\sin40°